Please help I will mark brainly

What is the distance between the points (-1,-9) and (4,-2)?

Crank

$\sqrt{74}$ , or 8.60

Just use the distance formula

4 0

3 years ago

Ted has $3.90. He has only dimes and quarters. He has 27 coins altogether. How many of each does he have

mestny [16]

Answer: 8 quarters 19 dimes

Step-by-step explanation:

we already know 8+19=27, if 4 quarters equals a dollar then you use 4 quarters each to create 2 dollars.

then, use 10 dimes to create another dollar summing it up to $3. now all we have to do is make 90 cents, and we've already used 18 coins.

if we used another 9 dimes, that would equal 90 cents.

in conclusion, 4+4= 8

8+10= 18

18+9= 27, and having $3.90

6 0

3 years ago

The amount people pay for cable service varies quite a bit but the mean monthly fee is $142 and the standard deviation is $29. t

zhuklara [117]

Answer:

a) By the Central Limit Theorem, the mean is $142 and the standard deviation is $0.7488.

b) By the Central Limit Theorem, approximately normal.

c) 0.0901 = 9.01% probability that the average cable service paid by the sample of cable service customers will exceed $143

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 $s = \frac{\sigma}{\sqrt{n}}$ .