Which expression is a factor of 10x^2+11x+3

What’s the answer to this? by the way you have to estimate

Furkat [3]

the answer is 2 if you estimate to the tenths place

7 0

3 years ago

Please find question in photo below. I’m revising for a test I have in a few days, and I can’t figure this out. Please help!

Temka [501]

$\large{\rm{\underline{\underline{Answer:}}}}$

a) n² + 6

~Here n is the number of term. So, put n = 1,2,3,4.... to get the term.

If n = 1, first term = 1² + 6 = 7
n = 2, second term = 10
n = 3, third temm = 15
n = 4, fourth term = 22
n = 10, tenth term = 106

b) 3n²

If n = 1, first term = 3
n = 2, second term = 12
n = 3, third term = 27
n = 4, fourth term = 48
n = 10, tenth term = 300

3 0

3 years ago

A statistician calculates that 8% of Americans own a Rolls Royce. If the statistician is right, what is the probability that the

hichkok12 [17]

Answer:

0.007 = 0.7% probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$ .

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

A statistician calculates that 8% of Americans own a Rolls Royce.

This means that $p = 0.08$

Sample of 595:

This means that $n = 595$

Mean and standard deviation:

$\mu = p = 0.08$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.08*0.92}{595}} = 0.0111$

What is the probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%?

Proportion above 8% + 3% = 11% or below 8% - 3% = 5%. Since the normal distribution is symmetric, these probabilities are equal, and so we find one of them and multiply by 2.

Probability the proportion is less than 5%:

P-value of Z when X = 0.05. So

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

By the Central Limit Theorem

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

$Z = \frac{0.05 - 0.08}{0.0111}$

$Z = -2.7$

$Z = -2.7$ has a p-value of 0.0035

2*0.0035 = 0.0070

0.007 = 0.7% probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%

8 0

3 years ago

When Elyse moved last weekend, the circular glass top for an end table broke. To replace it, she bought a circular glass top wit

spin [16.1K]

Answer:

D

Step-by-step explanation:

hmu for an explanation

4 0

3 years ago

Read 2 more answers

.What is the value of the expression below when x=9? Enter the answer as NUMBER only? 1/3 x^2+4

belka [17]

Step-by-step explanation:

1 *9^2 + 4

3

1 * 81 + 4

3

1 * 85

3

85

3

=28

5 0

3 years ago