All categories

3 years ago

Which of these is an example of a continuous random variable?

1 answer:

4 0

B.Number of eye blinks in a minute.
We do not have a set amount of blinks per minute.
A.Inccorect,a touchdown is 6 points
C.Speed of a race car rises at a constant rate depending on the car,speed e.t.c
D.Dogs get placed in a kennel and have a set ammount.

You might be interested in

Express 2/3 : 4/5 in lowest form

Mashcka [7]

Answer:

2/3 : 4/5 or 0.67 : 0.8

Step-by-step explanation:

it's already in its lowest form.

8 0

3 years ago

Use mapping notation to describe a translation up 8 units.

vitfil [10]

<span>Describe the translation in words: (x, y) (x - 3, y + 7) =3 units to the left, 7 units up

Describe the translation as an ordered pair: 5 units to the right, 4 units down.
=(x, y)(x + 5, y – 4)

Describe the translation in words: (x, y) (x + 6, y – 2) =6 units to the right, 2 units down

Describe the translation as an ordered pair: 1 unit to the left, 8 units down
=(x, y) (x – 1, y – 8)
</span>

6 0

4 years ago

Read 2 more answers

Arnoldo needs to write this system in slope-intercept form. Which shows how he could do that? 3 x minus 2 y = 6. 0.4 (20 y + 15)

kotegsom [21]

Answer:

y = three-halves x minus 3

Step-by-step explanation:

Subtract 3x from both sides of the original equation.

-2y = -3x +6

Divide by -2

y = 3/2x -3

4 0

3 years ago

Read 2 more answers

a triangle has the following side lengths: x-2,x, and 3x+1. another triangle has the following side lengths: 2x-5, x+4, 6x-7. of

choli [55]

Perimeter is the addition of all the sides
x-2+x+3x+1=2x-5+x+4+6x-7; since their perimeters are equal
5x-1=9x-8
collect the like terms
-1+8=9x-5x
7=4x
x=7/4
x=1.75

7 0

3 years ago

Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp

exis [7]

Answer:

$H_0: \, \mu = 2$ .

$H_1:\, \mu > 2$ .

Test statistics: $z \approx 2.582$ .

Critical value: $z_{1 - 0.025} \approx 1.960$ .

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean $\mu$ is greater than $2$ . This claim should be reflected in the alternative hypothesis:

$H_1:\, \mu > 2$ .

The corresponding null hypothesis would be:

$H_0:\, \mu = 2$ .

In this setup, the null hypothesis $H_0:\, \mu = 2$ suggests that $\mu_0 = 2$ should be the true population mean of GPA.

However, the alternative hypothesis $H_1:\, \mu > 2$ does not agree; this hypothesis suggests that the real population mean should be greater than $\mu_0= 2$ .

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed $\mu_0 = 2$ (i.e., the null hypothesis is true.) How likely would the sample (sample mean $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ ) be observed in this hypothetical population?

Let $\sigma$ denote the population standard deviation.

Given the large sample size $n = 60$ , the population standard deviation should be approximately equal to that of the sample:

$\sigma \approx s = 0.06$ .

Also because of the large sample size, the central limit theorem implies that $Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}}$ should be close to a standard normal random variable. Use a $Z$ -test.

Given the observation of $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ :

$\begin{aligned}z_\text{observed}&= \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} \\ &\approx \frac{\overline{X} - \mu_0}{s / \sqrt{n}} = \frac{2.02 - 2}{0.06 / \sqrt{60}} \approx 2.582\end{aligned}$ .

Because the alternative hypothesis suggests that the population mean is greater than $\mu_0 = 2$ , the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed $z$ -test. The question requested a significant level of $0.025$ . Therefore, the critical value $z_{1 - 0.025}$ should ensure that $P( Z > z_{1 - 0.025}) = 0.025$ .

Look up an inverse $Z$ table. The $z_{1 - 0.025}$ that meets this requirement is $z_{1 - 0.025} \approx 1.960$ .

The $z$ -value observed from the sample is $z_\text{observed}\approx 2.582$ , which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this $0.025$ confidence level in favor of the alternative hypothesis.

3 0

3 years ago