Combine the like terms to create an equivalent expression. 7z + 15 + 2

Answers

Answer 1

Answer:

Answer: 7z + 17

Step-by-step explanation: add 15 and 2

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What is the sum of two solutions of the quadratic equation ax^2+bx+c=0

Answers

if we were to use the quadratic formula, we would get that the roots are

$(-b\pm√(b^2-4ac))/(2a)$

or that the 2 seperate roots are

$(-b+√(b^2-4ac))/(2a)$ and $(-b-√(b^2-4ac))/(2a)$

if we sum these, then the $√(b^2-4ac)$ bits will cancel and we wil be left with

$(-b-b)/(2a)$ or $(-2b)/(2a)$ or $(-b)/(a)$

the sum of the solutions is $(-b)/(a)$

Try this option:

According to the properties of the quadratic equation the sum of its roots is '-b', in other words x₁+x₂= -b

Answer: -b

What's the nearest integer and the nearest tenth

Answers

Answer:

nearest integer 27 by 4

Order the Angel's from least to greatest with numbers 21, 24, 17

Answers

17,21,24

is the answer form least to greatest

17,21,24 is the answer

Consider random samples of size 40 from a population with proportion 0.15. (a) Find the standard error of the distribution of sample proportions.
Round your answer for the standard error to three decimal places.
mean=______
standard error=_______
(b) Is the sample size large enough for the Central Limit Theorem to apply?
1. Yes
2. No

Answers

The standard error of the distribution of sample proportions is 0.056 and mean is 0.15.

Yes, the sample size is enough for the Central Limit Theorem to apply.

(a). Given that, size of sample, $n=40$

Proportion, $p=0.15$

In the distribution of sample proportions, mean $\mu=p$

and, standard error = $\sqrt{(p(1-p))/(n) }$

So, mean $\mu=0.15$

Standard error = $\sqrt{(0.15(1-0.15))/(40) }=0.056$

(b). The Central Limit Theorem applies if np > 5 .

$np=40*0.15=6>5$

Thus, the Central Limit Theorem is applied.

Learn more:

brainly.com/question/22233199

Answer:

a) The mean is 0.15 and the standard error is 0.056.

b) 1. Yes

Step-by-step explanation:

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 $s = (\sigma)/(√(n))$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For proportions p, in samples of size n, the mean is $\mu = p$ and the standard error is $s = \sqrt{(p(1-p))/(n)}$ . The Central Limit Theorem applies is np > 5 and np(1-p)>5.