3x2 + 4x – 15 = (x

PLEASE HELP LOTS OF POINTS!!!!!!!! TWO QUESTIONS!!!!!!!!!

zubka84 [21]

Answer: Number one would be 3x+2y+-14 and x+y=-4, second one is x=-3, y=7

Step-by-step explanation:

The first one is solved by inputting the x and y in each one and finding which one comes out true, the second on is solved by substitution. to find x you would subtract x in the first equation and make it y=4-x then input that equation in the y in the second equation.

6 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

I'm in algebra 1 please help i'm so lost

9966 [12]

3+xx=19

x2+3=19

x2+3−3=19

−3x2=16

x=±√16

x = 4‌‌ or ‌‌x = −4

6 0

3 years ago

Multiply the number the number of minutes mr frotout spent on long distance calls by .45 the long distance rate then add 33 the

lara [203]

Answer:

0.45m + 33

Step-by-step explanation:

We are instructed to multiply the number the number of minutes Mr. Frotout spent on long distance calls by 0.45 the long distance rate then add 33 the flat monthly charge.

Assuming the number of long minutes spent on long distance calls by Mr. Frotout to be $m$ , we can express the given situation as follows: