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3 years ago

Estimate the difference. 9,136 - 3,572

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

7 0

ANSWER:
5,564
round it = 5600

olga2289 [7]3 years ago

6 0

Answer:

5500..?

Step-by-step explanation:

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the ratio of apples to oranges in a fruit basket is 12:4 if there are 7 oranges in the fruit basket then how many apples and ora

Ugo [173]

Just simple proportions:

12(apples) x(apples)
-------------- = --------------
4(oranges) 7(oranges)

12*7 = 84
84/4 = 21

There are 21 apples + 7 oranges = 28 apples and oranges!

7 0

3 years ago

Solve the system using elimination<br> 3x+y=27<br> -3x+4y=-42

padilas [110]

Hello :)

3x+y=27

-3x+4y= -42

+____________

(3x -3x) + (y+4y) = (27-42)

0 + 5y = -15

y = -3

Put "-3" for "y" in the system:

3x + y =27

3x -3=27

3x = 30

x = 10

Answer = x:10 y: -3

Have a nice day :)

8 0

3 years ago

A union of restaurant and foodservice workers would like to estimate the mean hourly wage, μ, of foodservice workers in the U.S.

pav-90 [236]

Answer:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be 95% confident that its estimate is within $0.35 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.15 .

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)

$\sigma=2.25$ represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ 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 =0.35 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (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: