I need help with probability help me answer this question correctly.

an 8 by 19 ft wall has three 4 by 5 ft windows.the area of the three windows is what percent of the area of the entire wall (inc

erica [24]

Answer:

39.5% (3 sf)

Step-by-step explanation:

Area of 1 window: 4×5 = 20 ft²

2 windows: 3×20 = 60ft²

Area of the wall: 8×19 = 152 ft³

Percentage of window area:

60/152 × 100 = 39.47368421%

= 39.5% (3 sf)

4 0

3 years ago

X4+3x-28 plz solve the question it's urgent plz

Margarita [4]

Answer:

x≈-2.433 ,x≈2.15

Step-by-step explanation:

5 0

3 years ago

I really need help with distributive property

docker41 [41]

Answer:

To “distribute” means to divide something or give a share or part of something. According to the distributive property, multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Two circles that have l as a common internal tangent.

KatRina [158]

Two circles that have l as a common internal tangent are B and C.

6 0

3 years ago

Read 2 more answers

A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is current

lapo4ka [179]

Using the z-distribution, we have that:

a)

The hypothesis test is:

$H_0: \mu = 7.4$

$H_1: \mu < 7.4$

The p-value is of 0.0668.

The critical value is $z^{\ast} = -1.28$

b) Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

Item a:

At the null hypothesis, it is tested if the mean is of 7.4 minutes, that is:

$H_0: \mu = 7.4$

At the alternative hypothesis, it is tested if the mean is lower than 7.4 minutes, that is:

$H_1: \mu < 7.4$

We have the standard deviation for the population, hence, the z-distribution is used.

The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

$\sigma$ is the standard deviation of the sample.

n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 7.1, \mu = 7.4, \sigma = 1.2, n = 36$ .

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{7.1 - 7.4}{\frac{1.2}{\sqrt{36}}}$

$z = -1.5$

Using a z-distribution calculator, the p-value is of 0.0668.

Also using a calculator, considering a left-tailed test, as we are testing if the mean is less than a value, the critical value with a significance level of 0.1 is of $z^{\ast} = -1.28$ .

Item b:

Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

You can learn more about the use of the z-distribution to test an hypothesis at brainly.com/question/16313918

5 0

2 years ago