All categories

3 years ago

Line Q is represented by the following equation: 2x + y = 11

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

8 0

Solve each eq simultaneously, because when u do when u get x or y value same as (3,5) that equation will be ur answer

You might be interested in

Help me real answers

Slav-nsk [51]

That is a vertical stretch by 2

8 0

2 years ago

The average production cost for major movies is 57 million dollars and the standard deviation is 22 million dollars. Assume the

Degger [83]

Using the normal distribution, we have that:

The distribution of X is $X \approx (57,22)$ .

The distribution of $\mathbf{\bar{X}}$ is $\bar{X} \approx (57, 5.3358)$ .

0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem, the parameters are given as follows:

$\mu = 57, \sigma = 22, n = 17, s = \frac{22}{\sqrt{17}} = 5.3358$

Hence:

The distribution of X is $X \approx (57,22)$ .

The distribution of $\mathbf{\bar{X}}$ is $\bar{X} \approx (57, 5.3358)$ .

The probabilities are the <u>p-value of Z when X = 58 subtracted by the p-value of Z when X = 55</u>, hence, for a single movie:

X = 58:

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

$Z = \frac{58 - 57}{22}$

Z = 0.05.

Z = 0.05 has a p-value of 0.5199.

X = 55:

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

$Z = \frac{55 - 57}{22}$

Z = -0.1.

Z = -0.1 has a p-value of 0.4602.

0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

For the sample of 17 movies, we have that:

X = 58:

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

$Z = \frac{58 - 57}{5.3358}$

Z = 0.19.

Z = 0.19 has a p-value of 0.5753.

X = 55:

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

$Z = \frac{55 - 57}{5.3358}$

Z = -0.38.

Z = -0.38 has a p-value of 0.3520.

0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

8 0

1 year ago

Simplify square root of 2 over cube root of 2.

Fed [463]

Answer:

2 to the power of one sixth

Step-by-step explanation:

Assuming you don't already know this, any type of root can be expressed as an exponent. Generally speaking:

$\sqrt[n]{x} = {x}^{ \frac{1}{n} }$

So you can rewrite the given fraction as

$\frac{ {2}^{ \frac{1}{2} } }{ {2}^{ \frac{1}{3} } }$

and then reduce as you normally would. That is, if the bases of the numerator and denominator are the same, then you can subtract the denominator's exponent from the numerator's exponent like so:

$\frac{ {2}^{ \frac{1}{2} } }{ {2}^{ \frac{1}{3} } } = {2}^{ \frac{1}{2} - \frac{1}{3} }$