Describe the transformation from f(x)=x for m(x)=(x-5) Ask for details

Mathematics

1 answer:

Likurg_2 [28]3 years ago

5 0

Answer:

m(x)=f(x)-5

the line formed by f(x) when transformed into m(x) shifts down 5 units on the (x,y)-plane

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(2/5pq^4)^2 • (-27p^5q)

miv72 [106K]

Answer: - 108q^8 p^2+5q/25

Step-by-step explanation:

5 0

2 years ago

Enter the value, as a mixed number in simplest form, in the box. |2 1/3| = ?

Fittoniya [83]

Unclear of the question, but the value is already as a mixed fraction. The value of absolute value is just the distance from zero. So, |-4|=4 or |6|=6
$|2 \frac{1}{3} | = 2 \frac{1}{3}$

7 0

3 years ago

According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe

Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

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

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

It 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, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .