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Vitek1552 [10]
3 years ago
11

Que es el slope o mas bien dicho que significa​

Mathematics
1 answer:
vekshin13 years ago
3 0
The slope is the rate of the line or the ratio between two points in a line
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a randon sample of 16 bookcases in one company have a mean height of 67.5 inches and a standard deviation of 2.1 inches. Constru
Lynna [10]

Answer:

For 99% of confidence interval is 67.5±1.3524

Step-by-step explanation:

Given:

Mean height =67.5 inches

Standard deviation:2.1 inches

Z at 99%.

No of samples 16.

To find:

confidence interval

Solution:

We have formula for confidence interval,

=mean ±Z*{standard deviation/sqrt(no.of observation)}

Now

Z=99%

has  standard  value as ,

Z=2.576

Confidence interval= mean±Z{standard deviation/sqrt(No. of samples)}

=67.5±2.57{(2.1/sqrt(16)}

=67.5±2.576(2.1/4)

=67.5±1.3524

5 0
3 years ago
a math class has a test today there are 30 problems on the test the test has two types of problems multiple choice problems and
DanielleElmas [232]
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7 0
3 years ago
The area of a rectangle is represented by the expression 6x³y5. Which of the following could be the
oee [108]

The length and width of the rectangle are: 2x^2y^2 , 3xy^3

<h3>What is the area of the rectangle?</h3>

The area of the rectangle is the product of the length and width of a given rectangle.

The area of the rectangle = length × Width

The area of a rectangle is represented by the expression 6x^3y^5.

The area of the rectangle = 2x^2y^2 \times 3xy^3

                                          = 6x^3y^5

So, the length and width of the rectangle are: 2x^2y^2 , 3xy^3

Learn more about the area;

brainly.com/question/1658516

#SPJ1

7 0
1 year ago
HELP ASAP FOR BRAINLIEST: The mean score for a standardized test is 1700 points. The results are normally distributed with a sta
WITCHER [35]

Answer:

Step-by-step explanation:

Since the results for the standardized test are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = test reults

µ = mean score

σ = standard deviation

From the information given,

µ = 1700 points

σ = 75 points

We want to the probability that a student will score more than 1700 points. This is expressed as

P(x > 1700) = 1 - P(x ≤ 1700)

For x = 1700,

z = (1700 - 1700)/75 = 0/75 = 0

Looking at the normal distribution table, the probability corresponding to the z score is 0.5

P(x > 1700) = 1 - 0.5 = 0.5

8 0
3 years ago
You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample
Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma=160)  

And the distribution for \bar X is:

\bar X \sim N(\mu, \frac{160}{\sqrt{n}})  

We know that the margin of error for a confidence interval is given by:  

Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)  

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.95=0.05 and \alpha/2=0.025  

Using the normal standard table, excel or a calculator we see that:  

z_{\alpha/2}=\pm 1.96  

If we solve for n from formula (1) we got:  

\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}  

n=(\frac{z_{\alpha/2} \sigma}{Me})^2  

And we have everything to replace into the formula:  

n=(\frac{1.96(160)}{78.4})^2 =16  

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

n=(\frac{z_{\alpha/2} \sigma}{Me})^2  

And now we can replace the new value of Me and see what we got, like this:

n=(\frac{1.96*160}{60})^2 =27.32

And if we round up the answer we see that the value of n to ensure the margin of error required Me=\pm 60 $ is n=28.    

5 0
2 years ago
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