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sweet [91]
3 years ago
14

A rectangular field is two times as long as it is wide. If the perimeter of the field is 450 feet, what are the dimensions of th

e field? What would the equation be and the width and the length of the field
Mathematics
1 answer:
lys-0071 [83]3 years ago
5 0

Hi There,

----------------------------------------

We Know:

Total Perimeter = 450 feet

Length is 2 times the width.

----------------------------------------

Solution:

Length = 150

Width = 75

75 x 2 = 150

150 + 150 + 75 + 75 = 450

----------------------------------------

Answer:

Length = 150 feet.

Width = 75 feet.

----------------------------------------

Hope This Helps :)

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Answer:

C.

Step-by-step explanation:

The bottom-right most cell tells us that the total number of students that responded to the survey is 310 students.

To find the answer, we can go through each choice.

A. Females taking Geometry

Row 1 Column 2 tells us that 53 females are taking geometry. 53/310 is about 17%.

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Answer:

  Dimensions of original room = 12 x 12 feet.

Explanation:

 Let the size of old square room be a x a.

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  So,     ( a+4 ) x ( a + 6 ) = a x a + 144

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The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a s
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Answer:

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that \mu = 1262, \sigma = 118

a) Determine the 26th percentile for the number of chocolate chips in a bag. ​

This is X when Z has a p-value of 0.26, so X when Z = -0.643.

Z = \frac{X - \mu}{\sigma}

-0.643 = \frac{X - 1262}{118}

X - 1262 = -0.643*118

X = 1186

(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.

Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.

2.5th percentile:

X when Z has a p-value of 0.025, so X when Z = -1.96.

Z = \frac{X - \mu}{\sigma}

-1.96 = \frac{X - 1262}{118}

X - 1262 = -1.96*118

X = 1031

97.5th percentile:

X when Z has a p-value of 0.975, so X when Z = 1.96.

Z = \frac{X - \mu}{\sigma}

1.96 = \frac{X - 1262}{118}

X - 1262 = 1.96*118

X = 1493

Between 1031 and 1493.

​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Difference between the 75th percentile and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

Z = \frac{X - \mu}{\sigma}

-0.675 = \frac{X - 1262}{118}

X - 1262 = -0.675*118

X = 1182

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

Z = \frac{X - \mu}{\sigma}

0.675 = \frac{X - 1262}{118}

X - 1262 = 0.675*118

X = 1342

IQR:

1342 - 1182 = 160

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