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2 years ago

A line has a slope of 0 and passes through the point (10, 2). What is its equation in slope intercept form?

Mathematics

2 answers:

Temka [501]2 years ago

6 0

Answer:

y = 10

Step-by-step explanation:

y = m x + C, where m is the slope and c is the intercept on the y axis. Plugging in the values of m and C, we get: y = 0 x + 10 = 0 + 10 = 10. Therefore, y = 10.

AleksAgata [21]2 years ago

3 0

I wrote the answer but realized I was wrong

Step-by-step

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Find the length of YA if A is between Y and Z; YA = 4x; AZ=16x and YZ = 200

olganol [36]

Answer:

200

Step-by-step explanation:

ya=yz

6 0

2 years ago

The difference of a number and five divided by 10

Hunter-Best [27]

Answer:

(n-5) ÷ 10

Step-by-step explanation:

7 0

3 years ago

I'm not sure what this means, please help me I'll give you 10 points ;)

Karo-lina-s [1.5K]

Answer:

So all the possible solutions are:

$\bold{523,~400,~379~and,~378.6}$

Solution Steps:

First you need to solve the real inequality to understand how to find the rest of the possible equations.

Add 78 to both sides:

$78+78=$ Cancels Out

$300+78=378$

So now we know the real answer, but it ask for all possible answers.

(Means anything larger than 300 when you plug it into x - 78 > 300.)

Numbers that are greater than 378:

1.) $377-78=299 > 300$ (False)

2.) $523-78=445>300$ (True)

3.) $0-78=-78>300$ (False)

4.) $-62-78=-140>300$ (False)

5.) $400-78=322>300$ (True)

6.) $379-78=301>300$ (True)

7.) $222-78=144>300$ (False)

8.) $378.6-78=300.6>300$ (True)

______________________________

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5 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 p-value of Z when X = 58 subtracted by the p-value of Z when X = 55, 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

A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced.

zzz [600]

Answer:

$CI=(66.54,78.46)$

Step-by-step explanation:

Given : A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced. The mean yield of the 10 batches is 72.5% and the standard deviation is 5.8%. Assume the yields are independent and approximately normally distributed.

To find : A 99% confidence interval for the mean yield when the new catalyst is used ?

Solution :

Let X be the yield of the batches.

We have given, n=10 , $\bar{X}=72.5\%$ , s=5.8%