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belka [17]
3 years ago
5

How do I solve this please help no links

Mathematics
1 answer:
Hitman42 [59]3 years ago
4 0

9514 1404 393

Answer:

  $735

Step-by-step explanation:

We assume the cover needs to cover the pool only, not any extra area around the pool.

The cost is given per square foot, so to find the total cost, we need to know how many square feet the cover must cover. The cost will be found by multiplying the cost per square foot by the number of square feet.*

The dimensions of the pool are given as 35 ft by 70 ft. The pool area is the product of these dimensions.

  A = LW

  A = (70 ft)(35 ft) = 2450 ft²

Then the cost is ...

  C = ($0.30/ft²)(2450 ft²) = $735

The cost to buy a pool cover is $735.

__

* This is the same rule you use for any purchase. The total cost is the cost per item multiplied by the number of items. Here, each "item" is 1 square foot of pool cover.

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Several programs attempt to address the shortage of qualified teachers by placing uncertified instructors in schools with acute
ss7ja [257]

Answer:

We conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

Step-by-step explanation:

We are given that reading scores of the students of certified teachers averaged 35.62 points with standard deviation 9.31. The scores of students instructed by uncertified teachers had mean 32.48 points with standard deviation 9.43 points on the same test.

There were 44 students in each group.

Let \mu_1 = <em><u>mean scores with uncertified teachers.</u></em>

\mu_2 = <em><u>mean scores with certified teachers.</u></em>

So, Null Hypothesis, H_0 : \mu_1\geq \mu_2     {means that the mean scores with uncertified teachers is higher or equal as compared to certified teachers}

Alternate Hypothesis, H_A : \mu_1     {means that the mean scores with uncertified teachers is lower as compared to certified teachers}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

                          T.S. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~ t__n__1-_n__2-2

where, \bar X_1 = sample mean scores of students instructed by uncertified teachers = 32.48 points

\bar X_2 = sample mean scores of students instructed by certified teachers = 35.62 points

s_1 = sample standard deviation of scores of students instructed by uncertified teachers = 9.43 points

s_2 = sample standard deviation of scores of students instructed by certified teachers = 9.31 points

n_1 = sample of students under uncertified teachers = 44

n_2 = sample of students under certified teachers = 44

Also,  s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(44-1)\times 9.43^{2} +(44-1)\times 9.31^{2} }{44+44-2} } = 9.37

So, <u><em>the test statistics</em></u>  =  \frac{(32.48-35.62)-(0)}{9.37 \times \sqrt{\frac{1}{44} +\frac{1}{44} } }  ~ t_8_6

                                     =  -1.572

The value of t test statistics is -1.572.

Since, in the question we are not given the level of significance so we assume it to be 5%. <u>Now, at 5% significance level the t table gives critical values of -1.665 at 86 degree of freedom for left-tailed test.</u>

Since our test statistic is more than the critical values of t as -1.572 > -1.665, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

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Answer: 6 triangles

Step-by-step explanation:

hope this helps :)

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2 years ago
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