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3 years ago

For 31 hours of work, you are paid $240.25. How much would you receive for 40 hours?

Mathematics

2 answers:

CaHeK987 [17]3 years ago

7 0

310

240.25/31= $7.75 an hour
7.75 x 40=$310

skad [1K]3 years ago

6 0

310 because you divide 240.25 by 31 and get 7.75. Then you subtract 40-31 and get 9 and multiply 9x7.75 and get 69.75. Lastly you add 69.75 and 240.25 and get 310. Hope that helps.

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3 years ago

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Answer:

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Step-by-step explanation:

Given Marcellus drank 1/4 gallon of the limeade and Marcy drank 1/8 gallon of the limeade

Given that only 5/8 portion of the pitcher contains limeade

Total amount drunk = amount drunk by Marcellus + amount drunk by Marcy

Total amount drunk = $\frac{1}{4} +\frac{1}{8}$ = $\frac{3}{8}$

The amount of Limeade remained = The amount initially - the amount drunk

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3 years ago

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2 years ago

Two statistics teachers both believe that each has the smarter class. To put this to the test, they give the same final exam to

SVETLANKA909090 [29]

Answer:

We conclude that there is no difference between the two classes.

Step-by-step explanation:

We are given that two statistics teachers both believe that each has a smarter class.

A summary of the class sizes, class means, and standard deviations is given below:n1 = 47, x-bar1 = 84.4, s1 = 18n2 = 50, x-bar2 = 82.9, s2 = 17

Let $\mu_1$ = mean age of student cars.

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1=\mu_2$ {means that there is no difference in the two classes}

Alternate Hypothesis, $H_A$ : $\mu_1\neq \mu_2$ {means that there is a difference in the two classes}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(47-1)\times 18^{2}+(50-1)\times 17^{2} }{47+50-2} }$ = 17.491

So, the test statistics = $\frac{(84.4-82.9)-(0)}{17.491 \times \sqrt{\frac{1}{47}+\frac{1}{50} } }$ ~ $t_9_5$

= 0.422

The value of t-test statistics is 0.422.

Now, the P-value of the test statistics is given by;

P-value = P( $t_9_5$ > 0.422) = From the t table it is clear that the P-value will lie somewhere between 40% and 30%.

Since the P-value of our test statistics is way more than the level of significance of 0.04, so we have insufficient evidence to reject our null hypothesis as our test statistics will not fall in the rejection region.

Therefore, we conclude that there is no difference between the two classes.

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3 years ago