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2 years ago

Pls help me:) A-(4•7)^-4 B-1/28^2 C-7^-2/4^2 D-(4•7)^4

Mathematics

1 answer:

Sergio039 [100]2 years ago

8 0

Answer:

$\frac{1}{(28)^{2} }$

Step-by-step explanation:

Step 1;

Write expression

$4^{-2} . 7^{-2\\\frac{\frac{1}{4^{-2} } }{y\frac{1}{7^{-2} } } }$

$\frac{1}{4^{2} .7^{2} }$

$\frac{1}{(4x7)^{2} }$

$\frac{1}{(28)^{2} }$

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The following data were collected from 12 rain gauges in a park. Build a 95% CI for the mean rainfall at the park.

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Answer:

Critical values: $t_{\alpha/2}=-2.201$ $t_{1-\alpha/2}=2.201$

95% confidence interval would be given by (3.646;4.472)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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The data is:

4.65 3.89 2.73 4.35 3.80 4.86 4.33 4.37 4.76 4.05 3.05 3.87

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In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$

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On this case the average is $\bar X= 4.059$

=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

The sample standard deviation obtained was s=0.6503

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =12 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df=n-1=12-1=11$

We can find the critical values in excel using the following formulas:

"=T.INV(0.025,11)" for $t_{\alpha/2}=-2.201$

"=T.INV(1-0.025,11)" for $t_{1-\alpha/2}=2.201$

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

And we can use Excel to calculate the limits for the interval

Lower interval : "=4.059 -2.201*(0.6503/SQRT(12))" =3.646

Upper interval : "=4.059 +2.201*(0.6503/SQRT(12))" =4.472

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Answer:

Part a) $y=18.50x+100$

Part b) The total cost is $\$220.25$

Part c) see the explanation

Step-by-step explanation:

Part a) Write a linear equation to represent this situation

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Let

x ----> the number of hours

y ----> the total cost in dollars

we know that

The linear equation in slope intercept form is equal to

$y=mx+b$

where

m is the slope or the unit rate

b is the y-intercept or initial value (value of y when the value of x is equal to zero)

In this problem we have that

The unit rate or slope is equal to

$m=\$18.50\ per\ hour$

The y-intercept or initial value is equal to

$b=\$100$

substitute

$y=18.50x+100$

Part b) Use the equation to find the total cost to rent skis from 8:30 am to 3:00 pm

Remember that

3:00 pm=15:00

from 8:30 am to 3:00 pm, the total hours is equal to

15:00 -8:30=6:30=6.5 h

For x=6.5 h

substitute in the linear equation and solve for y

$y=18.50(6.5)+100\\y=\$220.25$

Part c) What does your answer mean in context of the problem?

we have that

y=220.25

That means -----> The total cost to rent skis for 6.5 hours is equal to $220.25

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