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4 years ago

I need help asapppppppp

Mathematics

1 answer:

arlik [135]4 years ago

3 0

Answer:

a) h(x)

b) q(x)

c) w(x)

Step-by-step explanation:

Finding the answers for this is easy. Take two values for x that shows a different Y-value for each function on the graph and place them in the equations suggested to see which equation match their Y-value on the graph.

Let's take x = -1 and x = 1.

a) h(x) = | -x + 3 |

| - (-1) + 3| = | 4 | = 4

| - 1 + 3 | = | 2 | = 2

h(x) passes through (-1,4) and (1,2)

b) q(x) = | x | + 3

|-1| + 3 = 4

| 1 | + 3 = 4

q(x) passes through (-1,4) and (1,4)

c) w(x) = | x+ 3 |

| -1 + 3 | = | 2 | = 2

| 1 + 3 | = | 4 | = 4

w(x) passes through (-1,2) and (1,4)

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3 years ago

3/19% of a quantity is equal to <br> blank of the quantity.

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3 years ago

Assume that a simple random sample has been selected from a normally distributed population. State the hypotheses, find the test

BaLLatris [955]

Solution

Hypotheses:

- The population mean is 132. In order to test the claim that the mean is 132, we should check for if the mean is not 132.

- Thus, the Hypotheses are:

$\begin{gathered} H_0:\mu=132 \\ H_1:\mu\ne132 \end{gathered}$

Test statistic:

- The test statistic has to be a t-statistic because the sample size (n) is less than 30.

- The formula for finding the t-statistic is:

$\begin{gathered} t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} \\ \\ where, \\ \bar{X}=\text{ Sample mean} \\ \mu=\text{ Population mean} \\ s=\text{ Standard deviation} \\ n=\text{ Sample size} \end{gathered}$

- Applying the formula, we have:

$\begin{gathered} t=\frac{137-132}{\frac{14.2}{\sqrt{20}}} \\ \\ t=\frac{5}{3.1752} \\ \\ t\approx1.5747 \end{gathered}$

Critical value:

- The critical value t-critical, is gotten by reading off the t-distribution table.

- For this, we need the degrees of freedom (df) which is gotten by the formula:

$\begin{gathered} df=n-1 \\ df=20-1=19 \end{gathered}$

- And then we also use the significance level of 0.1 and the fact that it is a two-tailed test to trace out the t-critical. (Note: significance level of 0.1 implies 10% significance level)

- This is done below:

- The critical value is 1.729

P-value:

- To find the p-value, we simply check the table for where the t-statistic falls.

- The t-statistic given is 1.5747. We simply check which values this falls between in the t-distribution table. It falls between 1.328 and 1.729. We can simply choose a value between 0.1 and 0.05 and multiply the result by 2 since it is a two-tailed test.

- However, we can also use a t-distribution calculator, we have:

- Thus, the p-value is 0.13183

Final Conclusion:

- The p-value is 0.13183, and comparing this to the significance level of 0.1, we can see that 0.13183 is outside the rejection region.

- Thus, the result is not significant and we fail to reject the null hypothesis