Answer:
B. 120 units³
Step-by-step explanation:
The area of the large white rectangle on the right represents the area of one of the larger surfaces of the prism. It appears to be (10 units) × (6 units), so has an area of 60 units². (You can count the squares there, if you like.)
That is adjacent to a gray area on the net that is 2 units wide, indicating that the prism is 2 units deep. Thus the volume is ...
(60 units²) × (2 units) = 120 units³
Using the normal distribution, we have that:
- The distribution of X is
.
- The distribution of
is
.
- 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.
- 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
In this problem, the parameters are given as follows:

Hence:
- The distribution of X is
.
- The distribution of
is
.
The probabilities are the <u>p-value of Z when X = 58 subtracted by the p-value of Z when X = 55</u>, hence, for a single movie:
X = 58:


Z = 0.05.
Z = 0.05 has a p-value of 0.5199.
X = 55:


Z = -0.1.
Z = -0.1 has a p-value of 0.4602.
0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.
For the sample of 17 movies, we have that:
X = 58:


Z = 0.19.
Z = 0.19 has a p-value of 0.5753.
X = 55:


Z = -0.38.
Z = -0.38 has a p-value of 0.3520.
0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.
More can be learned about the normal distribution at brainly.com/question/4079902
#SPJ1
Treat as an equality, except flip the sign direction when yo multiply or divide both sides by a negative number
first distribute
2(4+2x)<u>></u>5x+5
distribute
8+4x<u>></u>5x+5
minus 4x fro both sides
8<u>></u>x+5
minus 5 from both sides
3<u>></u>x
x<u><</u>3
x is less than or equal to 3