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3 years ago

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The temperature in Bloomington on Sunday was 32.6°C. On Monday, the temperature changed by -8.25°C. What was the temperature on

Monday? -40.85°C

1 answer:

Afina-wow [57]3 years ago

8 0

Subtract 8.25 from 32.6 (32.6-8.25) = 24.35

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A and b are positive integers. Find <img src="https://tex.z-dn.net/?f=-2a%2B3b" id="TexFormula1" title="-2a+3b" alt="-2a+3b" ali

lord [1]

A could be 2 while B could be 3, so -2a+3b turns into -4+9, which equals 5.

From what I know you can't really solve a a single equation with two-variables so it's just a matter of trial and error.

Just try plugging in a small number like 2 for a just to try it and you get 8b^2=72.

Divide everything by 8 to isolate b and you get that b^2=9.

Square root everything and you'll find that b=3. This is just one possible combination, I'm sure there are many more but this is obviously the one that was intended to be found.

Now that we know that a=2 and b=3 just plug them into the equation.

-2(2)+3(3)=?

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Step-by-step explanation:

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Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

<em>Let </em> $\mu_1$ <em> = true mean number of sales at location A.</em>

<em /> $\mu_2$ = <em>true mean number of sales at location B</em>

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

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 average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, <u><em>test statistics</em></u> = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

<em>Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

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3 years ago