Answer is
<span>C) y = -2x + 3 </span>

when x = 0, y = 3
when x = 1, y = -2(1) + 3 = -2 + 3 = 1
when x = 2, y = -2(2) + 3 = -4 + 3 = -1

solutions matched the table above

5 0

3 years ago

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Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean

LuckyWell [14K]

Answer:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

$df = n_1 +n_2 -2 = 10+15-2= 23$

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

$\bar X_1 = 1085$ sample mean for group 1

$\bar X_2 = 1034$ sample mean for group 2

$n_1 = 10$ sample size for group 1

$n_2 = 15$ sample size for group 2

$s_1 = 52$ sample deviation for group 1

$s_2 = 61$ sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: $\mu_1= \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

And the statistic is given by:

$t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$

And replacing we got:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 10+15-2= 23$

And the p value would be:

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

8 0

4 years ago

A function is created to represent the amount of money in your checking account. What restrictions would be made to the domain?

MatroZZZ [7]

The answer I got was D. (now I'm just typing to make 20 characters) blah blah blah

4 0

4 years ago

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