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kati45 [8]
2 years ago
11

Draw a frequency polygon for the following data:

Mathematics
1 answer:
const2013 [10]2 years ago
8 0

Answer:

See attachment

Step-by-step explanation:

Given

\begin{array}{ccccccc}{Marks} & {0-10} & {10-20} & {20-30} & {30-40} & {40-50}  & {50-60}\ \\ {Students} & {7} & {15} & {22} & {30} & {16} & {10} \ \end{array}

Required

The frequency polygon

We have:

\begin{array}{ccccccc}{Marks} & {0-10} & {10-20} & {20-30} & {30-40} & {40-50}  & {50-60}\ \\ {Students} & {7} & {15} & {22} & {30} & {16} & {10} \ \end{array}

First, we calculate the midpoint of each class

\begin{array}{ccccccc}{Midpoint} & {(0+10)/2} & {(10+20)/2} & {(20+30)/2} & {(30+40)/2} & {(40+50)/2}  & {(50+60)/2}\ \\ {Students} & {7} & {15} & {22} & {30} & {16} & {10} \ \end{array}

\begin{array}{ccccccc}{Midpoint} & {5} & {15} & {25} & {35} & {45}  & {55}\ \\ {Students} & {7} & {15} & {22} & {30} & {16} & {10} \ \end{array}

Lastly, we plot the midpoint against the frequency of students (see attachment)

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Answer:

42 - 24n

Step-by-step explanation:

(7-4n)  6

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4 0
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Jessie draws triangle ABC on a coordinate grid. The slope of line segment AB is Jessie then transforms triangle
balu736 [363]

Answer:

1) Supports Jessie's Claim

2) Does Not Support Jessi's Claim

3) Supports Jessie's Claim

4) Does Not Support Jessi's Claim

Step-by-step explanation:

The given transformations are;

1) Rotation of 180° around the origin

For a rotation of 180° around the origin, either clockwise or anti clockwise, for a given coordinate of the preimage (x, y), the coordinate of the image is (-x, -y)

Therefore, whereby the slope of the preimage, given two points (0, 0) and (2, 2), = (2 - 0)/(2 - 0) = 1

For the image with the points (0, 0) and (-2, -2), we have;

(-2 - 0)/(-2 - 0) = 1

Therefore, the slope of the preimage and the image are equal

Therefore, supports Jessie's Claim

2) For a reflection across the line y = 2, we have

We note that the line y = 2 is parallel to the x-axis

For a reflection across the x-axis, for a preimage (x, y), we have the coordinates of the image (x, -y)

However for the reflection across the line y = 2, we have;

For a preimage, (x, y), the coordinate of the image is (x, -y+4)

Given two points, of the preimage (0, 0) and (2, 2), we have the image given as (0, 4) and (2, -2 + 4) = (2, 2);

The slope of the preimage is (2 - 0)/(2 - 0) = 1

The slope of the image is (2 - 4)/(2 - 0) = -1

The slope of the line of the preimage and the image are different

Therefore, does Not Support Jessi's Claim

3) For a translation up 1.25 units, we note that the difference in the y and x values of the coordinates of the preimage and the image will be equal when finding the slope, and therefore, the slope of the figure of the preimage and the slope of the figure of the image will be equal

Therefore, supports Jessie's Claim

4) For a reflection across the x-axis, a point on the preimage, with coordinates (x, y) will form a point on the image with coordinates (x, - y)

For a preimage with points (0, 0) and (2, 2), we have the image as (0, 0) and (2, -2)

The slope of the preimage is (2 - 0)/(2 - 0) = 1

The slope of the image is (-2 - 0)/(2 - 0) = -1

The slope of the line of the preimage and the image are different

Therefore, does Not Support Jessi's Claim

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3 years ago
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neonofarm [45]
When you move a number to the opposite side of the equation, in this occasion it will be 23- it changes to the opposite value. So us 23 is a positive value you can move it and change it to a negative. Or in other words suntract -23 from all sides of the equation, which will be equal to: 
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Find the indicated probability. round to three decimal places. a test consists of 10
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To answer this problem, we use the binomial distribution formula for probability:

P (x) = [n! / (n-x)! x!] p^x q^(n-x)

Where,

n = the total number of test questions = 10

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q = probability of failure = 0.5

Given the formula, let us calculate for the probabilities that the student will get at least 6 correct questions by guessing.

P (6) = [10! / (4)! 6!] (0.5)^6 0.5^(4) = 0.205078

P (7) = [10! / (3)! 7!] (0.5)^7 0.5^(3) = 0.117188

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P (9) = [10! / (1)! 9!] (0.5)^9 0.5^(1) = 0.009766

P (10) = [10! / (0)! 10!] (0.5)^10 0.5^(0) = 0.000977

Total Probability = 0.376953 = 0.38 = 38%

<span>There is a 38% chance the student will pass.</span>

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