1/3y=3 Find the value of y

Which of the following does not describe a control group? O A. a group of patients given a new drug B. a group of patients given

Veronika [31]

Answer:

the answer = C (may be but not sure)

5 0

3 years ago

Read 2 more answers

Help please will award brainliest

laiz [17]

Answer:

B)

Step-by-step explanation:

For the function $y=-5x$ , all coordinates this line passes through should make this inequality true.

$\begin{case}y=5(-7)=35\\y=-2(-5)=10\\y=2(-5)=-10\end{case}$

This corresponds with table B.

7 0

3 years ago

A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su

Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let $\mu_1$ = mean sales per market in Region 1.

$\mu_2$ = mean sales per market in Region 2.

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, $H_A$ : > $\mu_1-\mu_2\neq$ 0 {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about 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 sales in Region 1 = 84

$\bar X_2$ = sample mean sales in Region 2 = 78.3

$s_1$ = sample standard deviation of sales in Region 1 = 6.6

$s_2$ = sample standard deviation of sales in Region 2 = 8.5

$n_1$ = sample of supermarkets from Region 1 = 12

$n_2$ = sample of supermarkets from Region 2 = 17

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times 8.5^{2} }{12+17-2} }$ = 7.782

So, the test statistics = $\frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }$ ~ $t_2_7$

= 1.943

The value of t-test statistics is 1.943.

Now, at a 0.02 level of significance, the t table gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0

3 years ago

What are the x and y-intercepts of the line described by the equation? 3x−9y=10.8

marysya [2.9K]

$3x-9y=10.8\\\\x-intercept\ for\ y=0:\\\\3x-9(0)=10.8\\\\3x-0=10.8\\\\3x=10.8\qquad\text{divide both sides by 3}\\\\\boxed{x=3.6\to(3.6,\ 0)}\\\\y-intercept\ for\ x=0:\\\\3(0)-9y=10.8\\\\0-9y=10.8\\\\-9y=10.8\qquad\text{divide both sides by (-9)}\\\\\boxed{y=-1.2\to(0,\ -1.2)}$

3 0

3 years ago

Jennifer brought $ 24.75 to the baseball game. She spent $ 12.45 on drinks and snacks. How much money does she have left over?

Arturiano [62]

If you do 24.75-12.45 you get 12.30. So there you go!

5 0

3 years ago