Identify the pair of alternate exterior angles.

The area of a rectangle is 320 in^2. The ratio of the length to the width is 5:4. Find the length and the width.

Leviafan [203]

Answer:

length: 20 in
width: 16 in

Step-by-step explanation:

Let w represent the width of the rectangle. Then the length is 5/4w, and the area is ...

A = LW

320 in² = (5/4w)(w)

256 in² = w² . . . . . . . . . multiply by 4/5

16 in = w . . . . . . . . . . . . take the square root

5/4w = 5/4(16 in) = 20 in . . . . the length

The length of the rectangle is 20 inches; its width is 16 inches.

8 0

2 years ago

Two tables are placed side by side. One table is 5 feet 11 inches wide, and the other is 3 feet 6 inches wide. Together, how wid

babunello [35]

9 feet 5 inches
you just add the wideness of each one together

7 0

2 years ago

Read 2 more answers

The Dow Jones Industrial Average has had a mean gain of 432 pear year with a standard deviation of 722. A random sample of 40 ye

trasher [3.6K]

Answer:

66.98% probability that the mean gain for the sample was between 250 and 500.

Step-by-step explanation:

To solve this problem, it is important to know the Normal probability distribution and the Central limit theorem.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .