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3 years ago

Given m || n, find the value of x and y

Mathematics

1 answer:

Oksi-84 [34.3K]3 years ago

7 0

Answer:

x=33 and y= 75 is the answer.

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7 - r/8 - 3d Variables r=32 and d= 2

satela [25.4K]

The required value if r = 32 and d = 2 is -3

Given the expression

$7-\frac{r}{8} - 3d$

Substituting the given variables

r = 32

d = 2

The expression becomes

$= 7-\frac{32}{8} - 3(2)\\= 7 - 4 - 6\\= 3 - 6\\= -3$

Hence the required value is -3

Learn more here: brainly.com/question/17499155

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2 years ago

What is the expression for "the difference of 3 and 5 times a number"

soldi70 [24.7K]

(3 - 5)x

Hope this helps! :)

8 0

3 years ago

Read 2 more answers

The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books check

Verdich [7]

Answer:

$n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11$

So the answer for this case would be n=11 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma =150$ represent the population standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The confidence interval for this case is given by: (740, 920)

We can find the estimate for the mean and we got:

$\bar X = \frac{740+920}{2} = 830$

and the margin of error is given by :

$ME = \frac{920-740}{2}= 90$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =90 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got: