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Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.1 Triangles existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.1 Triangles

Exercises

IDENTIFY

Label the triangle as acute, right, or obtuse.

Question 1.

Answer:
Acute Angle,
Explanation:
An angle which is measuring less than 90 degrees is called an acute angle.
This angle is smaller than the right angle (which is equal to 90 degrees).
For example, ∠30^o, ∠45^o, ∠60^o, 75^o, etc. are all acute angles.

Question 2.

Answer:
Obtuse Angle,
Explanation:
The definition of an obtuse angle in geometry states that,
an angle whose measure is greater than 90° and less than 180° is called an obtuse angle.

Question 3.

Answer:
Right Angle,
Explanation:
x° = 60°
A right angle is an angle of 90°.
When two rays intersect and form a 90˚ angle or are perpendicular to each other,
at the intersection, they are said to form a right angle.

Question 4.

Answer:
Right Angle,
Explanation:
A right angle is an angle of 90°.
When two rays intersect and form a 90˚ angle or are perpendicular to each other,
at the intersection, they are said to form a right angle.

Question 5.

Answer:
Acute Angle,
Explanation:
An angle which is measuring less than 90 degrees is called an acute angle.
This angle is smaller than the right angle (which is equal to 90 degrees).

Question 6.

Answer:
Obtuse angle,
Explanation:
The definition of an obtuse angle in geometry states that,
an angle whose measure is greater than 90° and less than 180° is called an obtuse angle.

Label the triangle as equilateral, isosceles, or scalene.

Question 7.

Answer:
Isosceles,
Explanation:
An isosceles triangle is a triangle with two equal sides.
In the figure above, the two equal sides have same length.
An Isosceles triangle therefore has both two equal sides and two equal angles.

Question 8.

Answer:
Equilateral,
Explanation:
An equilateral triangle is a triangle with all three sides of equal length.
Total sum of triangle = 180°.
Given 2 angles = 60° + 60° = 120°.
Third angle = 180° – 120° = 60°.
Hence, the given angle is Equilateral triangle.

Question 9.

Answer:
Scalene Triangle,
Explanation:
All angles of a scalene triangle are unequal, all are of different size.
A scalene triangle has no line of symmetry.
The angle opposite to the longest side would be the greatest angle and vice versa.
All sides of the given scalene triangle are unequal.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 25.2 Types of Angles existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 25.2 Types of Angles

Exercises

IDENTIFY

Question 1.
What is the measure of angle Q?

Answer:
50°

Explanation:
The sum of the two angles are 180 degrees.
So, ∠Q = 90° – 40° = 50°

Question 2.
What is the measure of angle Z?

Answer:
103°
Explanation:
The sum of two angles are 180°
∠A = 77°
So, ∠Z = 180° – 77° = 103°

Question 3.
Identify the supplementary angles in the figure below.

Answer:
∠T and ∠V; ∠T and ∠S; ∠S and ∠X; ∠X and ∠V
Explanation:
Two angles are called supplementary when their measures add up to 180 degrees.

Question 4.
The measure of ∠G is given. Determine the measure of the remaining angles.

Answer:
∠F = 120°; ∠E = 60°; ∠H = 120°
Explanation:
Two angles are called supplementary when their measures add up to 180 degrees.
∠G = 60°
So, ∠F = 180° – 60° = 120°
∠E = 180° – 120° = 60°
∠H is opposite of ∠F
So, ∠H = 120°

SOLVE

Question 1.
From this figure, give examples of two complementary, two supplementary, and two vertical angles.

Complementary _______________
Supplementary _____________
Vertical ______________
Answer:
Complementary angles are ∠BGC and ∠BGA; ∠FGE and ∠DGE.
Supplementary angles are ∠BCG and ∠BGF; ∠AGC and ∠AGF; ∠BGA and ∠AGE.
Vertical angles are ∠BGC and ∠FGE; ∠DGC and ∠AGF; ∠AGB and ∠DGE.
Explanation:
Two angles are called complementary when their measures add to 90 degrees.
Two angles are called supplementary when their measures add up to 180 degrees.
Vertical angles are angles that are opposite of each other when two lines cross.
Vertical angles are always congruent.

Question 2.
For the figure below, list all complementary and supplementary angles.

Complementary _______________
Supplementary _____________
Answer:
Complementary angles are ∠DFE and ∠CFD; ∠CFB and ∠AFB.
Supplementary angles are ∠AFB and ∠EFB; ∠AFC and ∠CFE; ∠DFE and ∠DFA.
Explanation:
Two angles are called complementary when their measures add to 90 degrees.
Two angles are called supplementary when their measures add up to 180 degrees.

Question 3.

Are angle DOB and angle DOC complementary? Explain.
Answer:
Yes, they form a right angle.
Explanation:
When we add two angles, there sum should be 90°,
∠DOB + ∠DOC = ∠COB
Given, ∠DOB = 40° and ∠COB = 90°
∠DOC = 90° – 40° = 50°
So, the given angles are complementary angles.

Question 4.
Identify the measure of the angles in this figure.

COD ______________
COF ______________
FOY ______________
YOX ______________
FOX ______________
Answer:
COD = 50°
COF = 90°
FOY = 40°
YOX = 50°
FOX = 90°

Explanation:
The measure of angles formed by the two rays at a common vertex.
Angles are measured in degrees.
When we add two angles, there sum should be 90°,
∠DOB + ∠DOC = ∠COB
Given, ∠DOB = 40° and ∠COB = 90°
∠DOC = 90° – 40° = 50°
∠COF = 90°
∠DOB = 40° and ∠DOC = 50°
∠FOY = 40° and ∠YOX = 90°
∠FOY + ∠YOX = ∠FOX
40° + 50° = 90°

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 25.1 Measuring Angles existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 25.1 Measuring Angles

Exercises

IDENTIFY

Label the angles as acute, obtuse, or right.

Question 1.

Answer:
Acute,
Explanation:
If two rays intersect at a vertex, forming an angle that is less than 90° is known as acute.

Question 2.

Answer:
Obtuse,
Explanation:
Any angle that is greater than 90° but less than 180° is known as obtuse angle.

Question 3.

Answer:
Right angle,
Explanation:
If the angle formed between two rays is exactly 90° then it is called a right angle.

Question 4.

Answer:
Acute,
Explanation:
If two rays intersect at a vertex, forming an angle that is less than 90° is known as acute.

Question 5.

Answer:
Obtuse,
Explanation:
Any angle that is greater than 90° but less than 180° is known as obtuse angle.

Question 6.

Answer:
Acute,
Explanation:
If two rays intersect at a vertex, forming an angle that is less than 90° is known as acute.

Question 7.

∠DEF = _____________
Answer:
Right angle,
Explanation:
If the angle formed between two rays is exactly 90° then it is called a right angle.

Question 8.

∠RST = _____________
Answer:
Obtuse,
Explanation:
Any angle that is greater than 90° but less than 180° is known as obtuse angle.

Question 9.

∠LMN = _____________
Answer:
Obtuse,
Explanation:
Any angle that is greater than 90° but less than 180° is known as obtuse angle.

Question 10.

∠QRS = ______________
Answer:
Acute,
Explanation:
If two rays intersect at a vertex, forming an angle that is less than 90° is known as acute.

Question 11.

∠ABC = ______________
Answer:
Acute,
Explanation:
If two rays intersect at a vertex, forming an angle that is less than 90° is known as acute.

Question 12.

∠GHI = ______________
Answer:
Obtuse,
Explanation:
Any angle that is greater than 90° but less than 180° is known as obtuse angle.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 24.2 Line Segments and Rays existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 24.2 Line Segments and Rays

Exercises

IDENTIFY

Question 1.
How many line segments are there if you connect all the vertexes in this figure? Some, but not all, have been connected for you.

Answer:
30 lines segments,
Explanation:

there are 5 line segments from point A to B,C,D,E AND F.
6 points and 5 line segments,
6 x 5 = 30
Total 30 line segments can be constructed to connect all the vertexes in the above figure.

Question 2.
Let A stand for the town Augusta, and B stand for the town Bar Harbor. Using these two letters, what would be the ray if you were traveling from Bar Harbor through Augusta?
Answer:
\(\overline{BA}\),
Explanation:
Point A stand for the town Augusta, and B stand for the town Bar Harbor.
Using these two letters a ray from Bar Harbor through Augusta can be drawn.

Question 3.
Name all the possible line segments in this figure.

Answer:
\(\overline{BO}\), \(\overline{OD}\), \(\overline{BD}\), \(\overline{AB}\),
\(\overline{AO}\), \(\overline{OC}\), \(\overline{CB}\), \(\overline{BC}\),
\(\overline{OB}\), \(\overline{DO}\), \(\overline{DB}\), \(\overline{BA}\), \(\overline{OA}\), \(\overline{CO}\).
Explanation:
A line segment is part of a line that has two endpoints and is finite in length.
A ray is a line segment that extends indefinitely in one direction.
The following are all possible line segments, which can be drawn from O to B and O to D and all…
\(\overline{BO}\), \(\overline{OD}\), \(\overline{BD}\), \(\overline{AB}\),
\(\overline{AO}\), \(\overline{OC}\), \(\overline{CB}\), \(\overline{BC}\),
\(\overline{OB}\), \(\overline{DO}\), \(\overline{DB}\), \(\overline{BA}\), \(\overline{OA}\), \(\overline{CO}\).

Question 4.
At the right is a drawing of an airport runway. Name all the possible rays that could be used to describe the different ways a plane could take off. For points, use the numbers at the end of the runways: 15, 24, 8 and 33. An example of one is ray 15 – 33. Name the other three.

Answer:
Ray 33-15, Ray 24-8, Ray 8-24, Ray 15-33
Explanation:

A Ray from DC and CD Ray 33-15, Ray 15-33
A Ray from AB and BA Ray 24-8, Ray 8-24

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 24.1 Points and Lines existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 24.1 Points and Lines

Exercises

IDENTIFY

Question 1.
List the points located in the figure below.

Answer:
A, B, C, D
Explanation:
The below figure shown the points located A, B, C and D are vertex.
A point where two or more line segments join together are known as vertex.

Question 2.
How many different points are located in the figure below?

Answer:
8 points
Explanation:
A point where two or more line segments join together are known as vertex.

8 points are located in the above figure.
Each point is one vertex of the above figure.

Question 3.
Do the figures for exercises 1 and 2 contain any lines? Why or why not?
Answer:
No, line must stretch infinitely in both directions.
Explanation:
The above solid figures given in 1 and 2 dose not contain any lines.

Question 4.
List all possible names for the line pictured below.

Answer:
\(\overleftrightarrow{XY}\), \(\overleftrightarrow{YX}\), \(\overleftrightarrow{ZX}\), \(\overleftrightarrow{XZ}\), \(\overleftrightarrow{ZY}\) and \(\overleftrightarrow{YZ}\),
Explanation:

The above figure is a line with two points on the line X and Y with a center point Z and pointed.
the possible names for the line pictured are written below,
\(\overleftrightarrow{XY}\), \(\overleftrightarrow{YX}\), \(\overleftrightarrow{ZX}\), \(\overleftrightarrow{XZ}\), \(\overleftrightarrow{ZY}\), \(\overleftrightarrow{YZ}\).

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 3.1 Dividing Whole Numbers existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 3.1 Dividing Whole Numbers

Exercises Divide and Round To the nearest Hundredth

Question 1.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 395 and divisor is 7. Divide 395 by 7 the quotient is equal to 56.427 and remainder is 4.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 7 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 56.43.

Question 2.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 950 and divisor is 21. Divide 950 by 21 the quotient is equal to 45.238 and remainder is 2.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 8 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 45.24.

Question 3.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 9475 and divisor is 12. Divide 9475 by 12 the quotient is equal to 789.583 and remainder is 4.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 3 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 789.58.

Question 4.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 18980 and divisor is 36. Divide 18980 by 36 the quotient is equal to 527.222 and remainder is 8.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 2 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 527.22.

Question 5.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 8317 and divisor is 13. Divide 8317 by 13 the quotient is equal to 639.769 and remainder is 3.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 9 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 639.77.

Question 6.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 2000 and divisor is 44. Divide 2000 by 44 the quotient is equal to 45.454 and remainder is 24.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 4 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 45.45.

Question 7.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 256 and divisor is 3. Divide 256 by 3 the quotient is equal to 85.333 and remainder is 1.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 3 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 85.33.

Question 8.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 801 and divisor is 22. Divide 801 by 22 the quotient is equal to 36.409 and remainder is 2.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 9 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 36.41.

Question 9.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 788 and divisor is 14. Divide 788 by 14 the quotient is equal to 56.285 and remainder is 10.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 5 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 56.29.

Question 10.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 15412 and divisor is 16. Divide 15412 by 16 the quotient is equal to 963.25 and remainder is 0.

Question 11.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 40477 and divisor is 41. Divide 40477 by 41 the quotient is equal to 987.243 and remainder is 37.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 3 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 987.24.

Question 12.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 3041 and divisor is 37. Divide 3041 by 37 the quotient is equal to 82.189 and remainder is 7.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 9 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 82.19.

Question 13.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 9503 and divisor is 45. Divide 9503 by 45 the quotient is equal to 211.177 and remainder is 35.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 7 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 211.18.

Question 14.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 477 and divisor is 13. Divide 477 by 13 the quotient is equal to 36.692 and remainder is 13.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 2 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 36.69.

Question 15.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 4655 and divisor is 32. Divide 4655 by 32 the quotient is equal to 145.468 and remainder is 24.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 8 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 145.47.

Question 16.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 13183 and divisor is 89. Divide 13183 by 89 the quotient is equal to 148.123 and remainder is 53.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 3 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 148.12.

Question 17.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 4442 and divisor is 7. Divide 4442 by 7 the quotient is equal to 634.571 and remainder is 3.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 1 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 634.57.

Question 18.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 10602 and divisor is 11. Divide 10602 by 11 the quotient is equal to 963.818 and remainder is 1.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 8 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 963.82.

Question 19.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 13501 and divisor is 14. Divide 13501 by 14 the quotient is equal to 964.357 and remainder is 2.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 7 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 964.36.

Question 20.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 1444 and divisor is 32. Divide 1444 by 32 the quotient is equal to 45.125 and remainder is 0.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 5 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 45.13.

Question 21.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 2983 and divisor is 38. Divide 2983 by 38 the quotient is equal to 78.5 and remainder is 0.

Question 22.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 257 and divisor is 3. Divide 257 by 3 the quotient is equal to 85.666 and remainder is 2.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 6 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 85.67.

Question 23.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 1039 and divisor is 11. Divide 1039 by 11 the quotient is equal to 94.454 and remainder is 6.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 4 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 94.45.

Question 24.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 1484 and divisor is 33. Divide 1484 by 33 the quotient is equal to 44.969 and remainder is 23.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 9 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 44.97.

Question 25.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 2677 and divisor is 41. Divide 2677 by 41 the quotient is equal to 65.292 and remainder is 28.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 2 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 65.29.

Question 26.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 733 and divisor is 8. Divide 733 by 8 the quotient is equal to 91.625 and remainder is 0.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 5 which is greater than or equal to 5. So, we need to add 1 to the hundredths place in order to round it to the nearest hundredth. Finally after rounding the quotient will be 91.63.

Question 27.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 166 and divisor is 12. Divide 166 by 12 the quotient is equal to 13.833 and remainder is 4.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 3 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 13.83.

Question 28.

Answer:

Explanation:
Perform division operation on above two numbers. Here dividend is 11616 and divisor is 74. Divide 11616 by 74 the quotient is equal to 156.972 and remainder is 72.
We have to consider the thousandths place value when we are rounding quotient to the nearest hundredth.
We can see thousandths place value is 2 which is less than 5. So, we need to remove the digit. Finally after rounding the quotient will be 156.97.

Question 29.
A tailor can repair about 20 garments in a day. How many days will it take him to repair 300 garments?
Answer:
20 garments = 1 day
300 garments = ? days
(300 x 1)/20 = 15 days
To repair 300 garments a tailor can take 15 days.

Question 30.
Ms. Bailey is making information packets for all attendees at an education seminar. Each packet contains a total of 28 pages. If Ms. Bailey uses 1,400 sheets of paper, how many people plan to attend the education seminar?
Answer:
Given,
Ms. Bailey is making information packets for all attendees at an education seminar.
Each packet contains a total of 28 pages.
Ms. Bailey uses 1,400 sheets of paper
1400 ÷ 28 = 50

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 23.1 Calculating Probabilities existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 23.1 Calculating Probabilities

Exercises

CALCULATE

Question 1.
If there are 4 girls and 8 boys in a gym class, what is the probability of picking a girl as the person to lead the exercises?
Answer:
\(\frac{4}{12}\) = \(\frac{1}{3}\)
Explanation:
There is a formula to figure out probability (P)
It is the number of favorable out comes (f) divided by the total number of all possible outcomes (o).
In mathematical terms P = \(\frac{f}{o}\)
P = \(\frac{f}{o}\)
= \(\frac{4}{4+8}\)
= \(\frac{4}{12}\)
= \(\frac{1}{3}\)

Question 2.
In a deck of 52 cards, what is the probability of drawing a king from the deck?
Answer:
\(\frac{4}{52}\) = \(\frac{1}{13}\)
Explanation:
There is a formula to figure out probability (P)
It is the number of favorable out comes (f) divided by the total number of all possible outcomes (o).
In mathematical terms P = \(\frac{f}{o}\)
P = \(\frac{f}{o}\)
In a deck of 52 cards, there are 4 kings,
possible out comes 4 out of 52
= \(\frac{4}{52}\)
= \(\frac{1}{13}\)
Question 3.
There are 4 pizzas: 2 cheese, 1 vegetable and 1 everything. What is the probability that if you open a box it will be a cheese pizza?
Answer:
\(\frac{2}{4}\) = \(\frac{1}{2}\)
Explanation:
There is a formula to figure out probability (P)
It is the number of favorable out comes (f) divided by the total number of all possible outcomes (o).
In mathematical terms P = \(\frac{f}{o}\)
P = \(\frac{2}{4}\)
P = \(\frac{1}{2}\)

Question 4.
George is 1 of 5 white horses that is in a group of 15 horses in total. If you see a white horse, what is the probability that it is George?
Answer:
\(\frac{1}{5}\)
Explanation:
There is a formula to figure out probability (P)
It is the number of favorable out comes (f) divided by the total number of all possible outcomes (o).
In mathematical terms P = \(\frac{f}{o}\)
o = 5 white horse
f = George is one of white horse
P = \(\frac{1}{5}\)
Question 5.
You have change in your pocket: 5 quarters, 3 dimes, 2 nickels, and 4 pennies. If you pull a coin out of your pocket, what is the probability that it will be a dime?
Answer:
\(\frac{3}{14}\)
Explanation:
There is a formula to figure out probability (P)
It is the number of favorable out comes (f) divided by the total number of all possible outcomes (o).
In mathematical terms P = \(\frac{f}{o}\)
5 quarters, 3 dimes, 2 nickels, and 4 pennies total 14 coins
the probability that it will be a dime
P = \(\frac{f}{o}\)
P = \(\frac{3}{5 + 3 + 2 + 4}\)
P = \(\frac{3}{14}\)

Question 6.
There are 3 different prizes at the fair: 1 first prize of $100, 3 second prizes of $50, and 5 third prizes of $10. If there are 35 participants in the contest, what is the probability that you will win a prize?
Answer:
\(\frac{9}{35}\)
Explanation:
There is a formula to figure out probability (P)
It is the number of favorable out comes (f) divided by the total number of all possible outcomes (o).
In mathematical terms P = \(\frac{f}{o}\)
f = 1 + 3 + 5 = 9 prizes
o = 35 participants
P = \(\frac{f}{o}\)
= \(\frac{9}{35}\)

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 23.2 Probability Models existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 23.2 Probability Models

Exercises

SOLVE

Question 1.
Complete the probability model.

Answer:

Explanation:
A probability model is a chart or table that shows the different out comes possible for an event and their relative probabilities.

Question 2.
If Lynda plays one more game. what is the approximate probability that the game will end in a draw?
Answer:
0.59
Explanation:

Question 3.
If Lynda plays 7 more games and wins them all, how does that change her probability model?

Answer:

Explanation:
A probability model is a chart or table that shows the different out comes possible for an event and their relative probabilities.

Question 4.
What is the probability that a game will end in a draw now?
Answer:
0.55
Explanation:
A probability model is a chart or table that shows the different out comes possible for an event and their relative probabilities.

Question 5.
Based on the new model, how many times more likely is Lynda to win than to lose?
Answer:
Two times.
Explanation:
There is a formula to figure out probability (P)
It is the number of favorable out comes (f) divided by the total number of all possible outcomes (o).
In mathematical terms P
P = \(\frac{f}{o}\)
P = \(\frac{30}{97}\)
P = \(\frac{14}{97}\)
2 times more likely is Lynda to win than to lose.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 23.3 Probability of Compound Events existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 23.3 Probability of Compound Events

Exercises

CALCULATE

Question 1.
Zane rolls a six-sided die twice. What is the probability that he will roll a 3 and then a 4?
Answer:
\(\frac{1}{36}\)
Explanation:
There is a formula to figure out probability (P)
It is the number of favorable out comes (f) divided by the total number of all possible outcomes (o).
In mathematical terms P = \(\frac{f}{o}\)
P = \(\frac{f}{o}\)
= \(\frac{1}{36}\)

Question 2.
Marisa draws a coin from a box that contains 7 quarters, 5 dimes, 10 nickels, and 8 pennies. After looking at the coin, she puts it back and draws another. What is the probability that the coins she drew were a dime and then a penny?
Answer:
\(\frac{2}{45}\)
Explanation:
There is a formula to figure out probability (P)
It is the number of favorable out comes (f) divided by the total number of all possible outcomes (o).
In mathematical terms P = \(\frac{f}{o}\)
a box that contains 7 quarters, 5 dimes, 10 nickels, and 8 pennies
7 + 5 + 10 + 8 = 30
P = \(\frac{f}{o}\)
= \(\frac{2}{45}\)

Question 3.
Mrs. Castanon chose two students’ names from a hat (she chose the second name without replacing the first name). There are 30 students in her class. What is the probability that she chose John and then Nicholle?
Answer:
\(\frac{1}{870}\)
Explanation:
Probability of choosing John = \(\frac{1}{30}\)
Remaining names of students in the hat = 30 – 1 = 29
Probability of choosing Nicholle = \(\frac{1}{29}\)
Total probability of choosing Jhon and Nicholle = \(\frac{1}{30}\) x \(\frac{1}{29}\)
The probability that Castanon chose John and then Nicholle = \(\frac{1}{870}\)

Question 4.
Lori and Sergio love lollipops. At the same time, they each take one lollipop from a bag containing 8 cherry, 4 orange, and 7 grape lollipops. What is the probability that they both got an orange one?
Answer:
\(\frac{2}{57}\)
Explanation:
Total number of lollipops = 8 + 4 + 7 = 19
Number of orange lollipops = 4
number of orange lollipops taken = 2
Probability of Lori choosing an orange lollipop = \(\frac{4}{19}\)
Remaining number of orange lollipops in the bag = 4 – 1 = 3
Probability of Lori choosing an orange lollipop = \(\frac{3}{18}\)
the probability that they both got an orange lollipop = \(\frac{4}{19}\) x \(\frac{3}{18}\)
= \(\frac{2}{57}\)

Question 5.
Kalee and Nick are choosing two different movies to watch. In their collection, they have 10 sci-fi movies, 12 comedies, and 5 dramas. If Kalee chooses the first movie at random and then Nick chooses the second movie at random, what is the probability that they both choose sci-fi movies?
Answer:
\(\frac{5}{39}\)
Explanation:
Total number of movies = 10 + 12 + 5 = 27
Number of sci-fi movies = 10
Probability of choosing an sci-fi movies = \(\frac{10}{27}\)
Remaining number of sci-fi movies = 10 – 1 = 9
Probability of choosing a sci-fi movie = \(\frac{9}{26}\)
the probability that they both got an orange lollipop = \(\frac{10}{27}\) x \(\frac{9}{26}\)
= \(\frac{5}{39}\)

Question 6.
Amee wants to make a simulation to help her find the probability that there will be one girl and one boy from her class chosen for student council, if two students in the class are chosen at random. How can she design her simulation?
Answer:
Answer may vary.
Explanation:
Probability is about estimating or calculating how likely or ‘probable’ something is to happen.
Amee should use black squares of paper to represent the number of boys,
and white squares of paper to represent the number of girls.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 22.7 Mean Absolute Deviation existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 22.7 Mean Absolute Deviation

Exercises

CALCULATE

Sarah and Sadie recorded the numbers of points they scored in each of 5 basketball games.

Question 1.
What is Sarah’s average?
Answer:
4.6
Explanation:
Average = \(\frac{sum of the scores}{number of the scores}\)
= \(\frac{4 + 6 + 5 + 4 + 4}{5}\)
= \(\frac{23}{5}\) = 4.6

Question 2.
What is Sadie’s average?
Answer:
4
Explanation:
Average = \(\frac{sum of the scores}{number of the scores}\)
= \(\frac{3 + 7 + 2 + 6 + 2}{5}\)
= \(\frac{20}{5}\) = 4

Question 3.
What is the mean absolute deviation for Sarah’s scores?
Answer:
mean = 0.72
Explanation:
Sarah’s scores
Average = \(\frac{sum of the scores}{number of the scores}\)
= \(\frac{4 + 6 + 5 + 4 + 4}{5}\)
= \(\frac{23}{5}\) = 4.6
finding the difference between score and average score
I 4 – 4.6 I = 0.6
I 6 – 4.6 I = 1.4
I 5 – 4.6 I = 0.4
I 4 – 4.6 I = 0.6
I 4 – 4.6 I = 0.6
average of the differences =
= \(\frac{0.6 + 1.4 + 0.4 + 0.6 + 0.6}{5}\)
= \(\frac{3.6}{5}\) = 0.72
mean absolute deviation (MAD) = 0.72

Question 4.
What is the mean absolute deviation for Sadie’s scores?
Answer:
mean = 2
Explanation:
Sarah’s scores
Average = \(\frac{sum of the scores}{number of the scores}\)
= \(\frac{3 + 7 + 2 + 6 + 2}{5}\)
= \(\frac{20}{5}\) = 4
finding the difference between score and average score
I 3 – 4 I = 1
I 7 – 4 I = 3
I 2 – 4 I = 2
I 6 – 4 I = 2
I 2 – 4 I = 2
average of the differences,
= \(\frac{1 + 3 + 2 + 2 + 2}{5}\)
= \(\frac{10}{5}\) = 2
mean absolute deviation (MAD) = 2.

Question 5.
How many times greater is the mean absolute deviation of Sadie’s scores than for Sarah’s scores?
Answer:
1.28
Explanation:
the mean absolute deviation of Sadie’s score
mean absolute deviation (MAD) = 2
the mean absolute deviation of Sarah’s score
mean absolute deviation (MAD) = 0.72
The mean absolute deviation of Sadie’s scores than for Sarah’s scores difference,
= 2.0 – 0.72
= 1.28