Which number is the same distance from 0 as −14?

GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m

Harrizon [31]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

Let $\mu$ = average mg of mercury in compact fluorescent bulbs.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean mg of mercury = 3.59

$\sigma$ = population standard deviation = 0.18 mg

n = sample of bulbs = 25

So, test statistics = $\frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }$

= 2.50

The value of z test statistics is 2.50.

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

P-value = P(Z > 2.50) = 1 - P(Z $\leq$ 2.50)

= 1 - 0.9938 = 0.0062

Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the 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 the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0

3 years ago

The table of values represents an exponential function f(x). What is the average rate of change over the interval −2≤x≤2? Enter

Sergio [31]

Answer:

-1456

-864

4

-11

-5,25

-1,93

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

I need help! Will mark Brainliest for full answer!!!

Softa [21]

Answer:

Part a .............> x = 11

Part b .............> k = 57.2

Part c .............> y = 9.2

Explanation:

The three problems deal with inverse variation between two variables

An inverse variation relation between two variables means that when one of the variables increases, the other will decrease (and vice versa)

Mathematically, an inverse variation relation is represented as follows:

$y = \frac{k}{x}$

where x and y are the two variables and k is the constant of variation

Now, let's check the givens:

Part a:

We are given that y = 3 and k = 33

Substitute in the original relation and solve for x as follows:

$y = \frac{k}{x}\\ \\3 = \frac{33}{x}\\ \\x=\frac{33}{3}=11$

Part b:

We are given that y = 11 and x = 5.2

Substitute in the original relation and solve for k as follows:

$y=\frac{k}{x}\\ \\11=\frac{k}{5.2}\\ \\k=11*5.2=57.2$

Part c:

We are given that x=7.8 and k=72

Substitute in the original relation and solve for y as follows:

$y=\frac{k}{x}=\frac{72}{7.8}=9.2$ to the nearest tenth

Hope this helps :)

6 0

2 years ago

Tara started reading at 8:50 A.M. and stopped reading at 11:05 A.M. How long did Tara read?

Gwar [14]

She read 3 hours and 15 minutes.

Hope this helps!

5 0

2 years ago

Please tell me answers

Anna11 [10]

Answer:

no 8 is 90° and no 9 is two

Step-by-step explanation:

90° and two circles can pass through a given point

8 0

3 years ago