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svetoff [14.1K]
3 years ago
5

Please help me on question 4 and 5 I am really stuck thank you I would really appreciate it

Mathematics
1 answer:
STALIN [3.7K]3 years ago
4 0

Answer:

1. 5/4

2. 7

Step-by-step explanation:

1) Lets call the width as w

Therefore length would be:

w+4

To find the perimeter you use the formula:

2 (l+w)

Now substitute our values into this formula:

2 (w+4+w)

2( 2w+4)

4w+8

Now make this equal to 13:

4w +8 = 13

4w = 5

w = 5/4

2. In this question we will call length l

Therefore width would be:

l-5

Now we will do the steps we did above:

2 (l+l-5)

2 (2l-5)

4l -10

4l - 10 = 18

4l = 28

l = 7

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Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean
Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 50, \sigma = 6

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

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

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

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

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

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

Z = \frac{67 - 50}{6}

Z = 2.83

Z = 2.83 has a pvalue 0.9977

X = 60

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

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

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

Z = \frac{76 - 50}{6}

Z = 4.33

Z = 4.33 has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

-1.28 = \frac{X - 50}{6}

X - 50 = -1.28*6

X = 42.32

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

-1.28 = \frac{X - 50}{6}

X - 50 = -1.28*6

X = 42.32

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

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

1.28 = \frac{X - 50}{6}

X - 50 = 1.28*6

X = 57.68

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

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B) Yes this is a function

C) This is a function because the weekly salary is dependent on the number of sales 's'. Hence, total weekly salary is a function of sales.

D) Domain will be [ 0, ∞ ]

Step-by-step explanation:

Given :

Weekly salary = $1000

Commission = 5% of all her sales

Let the number of sales be 's'

thus,

A) Total weekly salary = $1000 + ( 0.05 × s )

B) Yes this is a function

C) This is a function because the weekly salary is dependent on the number of sales 's'. Hence, total weekly salary is a function of sales.

D) The domain here will be all the possible values of sales i.e 's'

Hence,

sales will always be positive

thus,

domain will be [ 0, ∞ ]

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