Need help on answering number 16.

What is the area of this composite figure?<br>A. 135 cm2<br>B. 207 cm2<br>C. 249 cm2<br>D. 279 cm2

mariarad [96]

Answer:

249 cm^2

Step-by-step explanation:

This problem becomes easier if we subdivide the figure, find the areas of the resulting figures and then sum them up.

Draw a vertical line straight down from the edge marked "4 cm" towards the edge marked "18 cm." The resulting rectangle on the left is 15.5 cm long and (18 - 7.5) cm wide, or 15.5 by 10.5 cm. Its area is 162.75 cm^2.

Next, find the area of the rectangle on the right of the line we drew. Its width is 7.5 cm and its height (15.5 - 4) cm, resulting in an area of 86.25 cm^2.

Last, add together these two subareas: combine 86.25 cm^2 and 162.75 cm^2. The total area of the composite figure is then 249 cm^2 (answer).

7 0

3 years ago

The Office of Student Services at a large western state university maintains information on the study habits of its full-time st

Vera_Pavlovna [14]

Answer:

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 20 hours, standard deviation of 6:

This means that $\mu = 20, \sigma = 6$

Sample of 150:

This means that $n = 150, s = \frac{6}{\sqrt{150}}$

What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours?

This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.5. So

X = 21

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{21 - 20}{\frac{6}{\sqrt{150}}}$

$Z = 2.04$

$Z = 2.04$ has a p-value of 0.9793

X = 19.5

$Z = \frac{X - \mu}{s}$

$Z = \frac{19.5 - 20}{\frac{6}{\sqrt{150}}}$

$Z = -1.02$

$Z = -1.02$ has a p-value of 0.1539

0.9793 - 0.1539 = 0.8254

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

3 0

3 years ago

What is 8640 minutes in days

Zolol [24]

Answer:

it is 6 days

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Pluto is looking into joining a gym. He has a budget of $68 per month. The gym he wants to join has fees based on the number of

Artist 52 [7]

7(10 visits) ≤ 68
70 ≤ 68
Not Viable

7(9 visits) ≤ 68
63 ≤ 68
Viable

The first solution is not viable because 10 visits will give a total cost of $70 per month, which is $2 over his monthly budget of $68.

The second solution is viable. He could visit the gym a maximum of 9 times per month for a total of $63 a month and that will be under his $68 monthly budget.

Hope this helps! :)

6 0

4 years ago

Find the area of the circle... Round to the nearest tenth and include units in your answer

Fantom [35]

Answer:

452.4 cm

Step-by-step explanation:

the formula for the area of a circle is A = pi r^2

4 0

3 years ago