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TEA [102]
1 year ago
13

A safety officer wants to prove that μ = the average speed of cars driven by a school is less than 25 mph. Suppose that a random

sample of 14 cars shows an average speed of 24.0 mph, with a sample standard deviation of 2.2 mph. Assume that the speeds of cars are normally distributed. What is the p-value?
Mathematics
1 answer:
Akimi4 [234]1 year ago
8 0

Answer:

t=\frac{24-25}{\frac{2.2}{\sqrt{14}}}=-1.70    

The degrees of freedom are given by:

df=n-1=14-1=13  

The p value for this case would be given by:

p_v =P(t_{(13)}  

Step-by-step explanation:

Information given

\bar X=24 represent the mean height for the sample  

s=2.2 represent the sample standard deviation

n=14 sample size  

\mu_o =25 represent the value that we want to test

t would represent the statistic

p_v represent the p value for the test

Hypothesis to verify

We want to cehck if the true mean is lees than 25 mph, the system of hypothesis would be:  

Null hypothesis:\mu \geq 25  

Alternative hypothesis:\mu < 25  

The statistic would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{24-25}{\frac{2.2}{\sqrt{14}}}=-1.70    

The degrees of freedom are given by:

df=n-1=14-1=13  

The p value for this case would be given by:

p_v =P(t_{(13)}  

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The function A(b) relates the area of a trapezoid with a given height of 10 and
Natali5045456 [20]

The function of the trapezoid area is:

A(x)=(B+b)*h/2

Where B and b are the bases and h is the height.

With the given data: h=10 B and b =7 and x (it may vary which one is bigger)

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So that function becomes:

A(x)=(7+x)*10/2

A(x)=(7+x)*5

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So if you want the inverse function, you have to operate to find x:

A(x)/5=7+x

A(x)/5-7=x

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So the new function is:

x(A)=A/5-7

6 0
2 years ago
There are 11 portable mini suites (a.k.a. cages) in a row at the Paws and Claws Holiday Pet Resort. They are neatly labeled with
BigorU [14]

Step-by-step explanation:

a) 7!

If there are no restrictions, answer is 7! as it is the permutation of all animals.

b) 4! x 3!

As cats are 6 and Dogs are 5, thus 1st and last must be cats in order to have alternate arrangements. Therefore the only choices are the order of the cats among  themselves and the order of the dogs among themselves. There are 4! permutations of the cats and 3! permutations of the dogs, so there are a total of 4! x 3! possible arrangements of the suites.

c) 3! x 5!

There are 3! possible arrangements of  the dogs among themselves. Now, if we consider the dogs as  one ”object” together, then we can think of arranging the 4 cats  together with this 1 additional object. There are 5! such arrangements possible, so there are a total of 3! · 5! possible arrangements of the suites.

d) 2 x 4! x 3!

As required that all the cats must be together and all the dogs must be together, either the cats are all  before the dogs or the dogs are all before the cats. There are two possible arrangements thus two times of both possibilities is the answer i.e.  2 x 4! x 3!

3 0
2 years ago
Which equation represents the line shown on the graph?
Mama L [17]

Hello there,

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8 0
1 year ago
A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si
Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

Area = 128

Let the dimension of the paper be x and y;

Such that:

Length = x

Width = y

So:

Area = x * y

Substitute 128 for Area

128 = x * y

Make x the subject

x = \frac{128}{y}

When 1 inch margin is at top and bottom

The length becomes:

Length = x + 1 + 1

Length = x + 2

When 2 inch margin is at both sides

The width becomes:

Width = y + 2 + 2

Width = y + 4

The New Area (A) is then calculated as:

A = (x + 2) * (y + 4)

Substitute \frac{128}{y} for x

A = (\frac{128}{y} + 2) * (y + 4)

Open Brackets

A = 128 + \frac{512}{y} + 2y + 8

Collect Like Terms

A = \frac{512}{y} + 2y + 8+128

A = \frac{512}{y} + 2y + 136

A= 512y^{-1} + 2y + 136

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

A' = -512y^{-2} + 2

Set

A' = 0

This gives:

0 = -512y^{-2} + 2

Collect Like Terms

512y^{-2} = 2

Multiply through by y^2

y^2 * 512y^{-2} = 2 * y^2

512 = 2y^2

Divide through by 2

256=y^2

Take square roots of both sides

\sqrt{256=y^2

16=y

y = 16

Recall that:

x = \frac{128}{y}

x = \frac{128}{16}

x = 8

Recall that the new dimensions are:

Length = x + 2

Width = y + 4

So:

Length = 8 + 2

Length = 10

Width = 16 + 4

Width = 20

To double-check;

Differentiate A'

A' = -512y^{-2} + 2

A" = -2 * -512y^{-3}

A" = 1024y^{-3}

A" = \frac{1024}{y^3}

The above value is:

A" = \frac{1024}{y^3} > 0

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

3 0
1 year ago
Ben Carson was loaned $125.00 to pay $6.00 interest for $100.00 every week ,after 3 weeks how much do he pay back to clear the d
Ghella [55]

Answer:

Total amount pay after 3 week = $147.5

Step-by-step explanation:

Given:

Amount of loan taken = $125

Amount interest per $100 = $6 per week

Find:

Total amount pay after 3 week

Computation:

Amount of interest per week = 125[6/100]

Amount of interest per week = $7.5 per week

Amount of interest in three week = 3 x Amount of interest per week

Amount of interest in three week = 3 x 7.5

Amount of interest in three week = $22.5

Total amount pay after 3 week = Amount of loan taken + Amount of interest in three week

Total amount pay after 3 week = 125 + 22.5

Total amount pay after 3 week = $147.5

8 0
1 year ago
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