Domain and range of: s={(-9,6),(5,6),(-3,0)}

A formula for 8 bottles of window cleaner calls for 6 cups of rubbing alcohol, 2.25 gallons of water, and 1.5 cups of ammonia. H

tamaranim1 [39]

240 quarts----------

7 0

3 years ago

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"Immediately after a ban on using hand-held cell phones while driving was implemented, compliance with the law was measured. A r

sergiy2304 [10]

Answer:

(a) Null Hypothesis, $H_0$ : $p_1-p_2=0$ or $p_1= p_2$

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq 0$ or $p_1\neq p_2$

(b) We conclude that there is a statistical difference in these two proportions measured initially and then one year later.

Step-by-step explanation:

We are given that a random sample of 1,250 drivers found that 98.9% were in compliance. A year after the implementation, compliance was again measured to see if compliance was the same (or not) as previously measured.

A different random sample of 1,100 drivers found 96.9% compliance."

Let $p_1$ = proportion of drivers that were in compliance initially

$p_2$ = proportion of drivers that were in compliance one year later

(a) Null Hypothesis, $H_0$ : $p_1-p_2=0$ or $p_1= p_2$ {means that there is not any statistical difference in these two proportions measured initially and then one year later}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq 0$ or $p_1\neq p_2$ {means that there is a statistical difference in these two proportions measured initially and then one year later}

The test statistics that will be used here is Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{ \frac{\hat p_1(1-\hat p_1)}{n_1} + \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of drivers in compliance initially = 98.9%

$\hat p_2$ = sample proportion of drivers in compliance one year later = 96.9%

$n_1$ = sample of drivers initially = 1,250

$n_2$ = sample of drivers one year later = 1,100

(b) So, the test statistics = $\frac{(0.989-0.969)-(0)}{\sqrt{ \frac{0.989(1-0.989)}{1,250} + \frac{0.969(1-0.969)}{1,100}} }$

= 3.33

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

P-value = P(Z > 3.33) = 1 - P(Z $\leq$ 3.33)

= 1 - 0.99957 = 0.00043

Since in the question we are not given with the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical values between -1.96 and 1.96 for two-tailed test.

Since our test statistics does not lies within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that there is a statistical difference in these two proportions measured initially and then one year later.

7 0

3 years ago

Use the original price and the markup to find the retail price.

Natasha_Volkova [10]

41 dollarsStep-by-step explanation:

4 0

3 years ago

Completely Factor 2n^2-23n+45=0

Anton [14]

Answer:

x=5/2, 9

Step-by-step explanation:

3 0

3 years ago

I attached an image. someone help me asap. i'm not the smartest person so i cont understand it :(

vampirchik [111]

Answer:

for the second column it is 12.5

for the third column it is 20%

for the fourth column it is 120%

for the fifth column it is 33%

for the sixth column it is 11.35

for the seventh column it is 900

for the 8th column it is 60%

for the final column it's 39

Step-by-step explanation:

7 0

2 years ago

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