An airplane drops a flare from a height of 1024 ft above a lake. Use the formulat = to find how many seconds it

(х-y)²-3(x-y) how do you factorise this equation?

Kazeer [188]

Answer:

(x-y)(x-y-3)

Step-by-step explanation:

(x-y)² - 3(x-y)

take (x-y) common,

(x-y)(x-y-3)

4 0

3 years ago

The length of a rectangle is 2 centimeters more than five times the width. The perimeter is 114 inches. Find the width and the l

gogolik [260]

1. Let L=length and W=width
2. L=5w+2
3. The perimeter can be solved as follows:
(5w+2) + (5w+2) + w + w = 114 in
12w + 4 = 114 in
12w = 110
w = 110/12
w = 9.17
L = 47.83

7 0

2 years ago

Majesty Video Production Inc. wants the mean length of its advertisements to be 26 seconds. Assume the distribution of ad length

Paladinen [302]

Answer:

a) By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.

b) s = 0.44

c) 0.84% of the sample means will be greater than 27.05 seconds

d) 98.46% of the sample means will be greater than 25.05 seconds

e) 97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds

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 normally distributed samples are 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 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(also called standard error) $s = \frac{\sigma}{\sqrt{n}}$ .