Help with these two please

Mathematics

2 answers:

kolbaska11 [484]3 years ago

5 0

The answer to the question is 5 -5

ad-work [718]3 years ago

3 0

1. 2x^2=50 -> x^2=25 -> sqrt(x^2) = sqrt(25) -> x = +/- 5

2. (c+9)^2 = 64 -> square root both sides ->
(a) c+9=8 -> c= c= 8-9 -> c=-1
(b) c+9= -8 -> c= -8-9 -> c= -17

You might be interested in

A veterinarian does research on the causes of enteroliths, stones that develop in the colon of horses. She suspects that feeding

dsp73

Answer:

On average, horses with and without enteroliths are fed the same amount of alfalfa per month

Step-by-step explanation:

Hello!

The veterinarian wants to research if eating alfalfa causes horses to have enteroliths.

The study variables are:

X₁: number of alfalfa flakes eaten over a month by a horse with enteroliths.

X₂: number of alfalfa flakes eaten over a month by a horse free of enteroliths.

She calculated a CI interval, with level 1 - α and the interval contained the cero.

With a significance level α, the hypotheses are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

To decide whether you should reject or not the null hypothesis using a Confidence Interval is the following:

If the CI contains the number stated in the null hypothesis, the decision is to not reject the null hypothesis.

If the CI doesn't contain the number stated in the null hypothesis, then the decision is to reject the null hypothesis.

Since the interval contains the cero, if 1 - α and α are complementary, the decision is to reject the null hypothesis.

This means that at a level of significance of α, you can conclude that there is no difference between the population means of the numbers of alfalfa flakes eaten by horses with or without enteroliths.

I hope it helps!

7 0

3 years ago

Suppose that the quarterly sales levels among health care information systems companies are approximately normally distributed w

cupoosta [38]

Answer:

The cutoff sales level is 10.7436 millions of dollars

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 12, \sigma = 1.2$