Mark and his friends ordered 2 pizzas. The same size

1 answer:

5 0

We know that

They cut the first pizza into 6 equal slice---> each slice is 1/6 of a whole pizza

They cut the second pizza into 4 equal slices---> each slice is 1/4 of a whole pizza

Each person plans to get 2 slices

case 1) one slide pizza 1 and one slide pizza 2
total pizza=(1/6)+(1/4)----> (4+6)/24----> 10/24----> 5/12 of a whole pizza

case 2) two slides pizza 1
total pizza=2*(1/6)-----> 2/6----> multiply by 2-----> (2/2)*2/6---> 4/12
so
4/12 < 5/12-------> Mark is correct

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Insert grouping symbols in 5 x 2 + 8 - 4 - 2 so that its value is 15

Arlecino [84]

5×2+8-4+1=15
cause
5×2=10
10+8=18
18-4=14
14+1=15

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I know I haven't been on this app for a long time but can ya'll help me

kramer

Answer:

2774

Step-by-step explanation:

start by multiplying 2 by 7, the answer is 14 put the 1 above the 8 in 1387 then put the 4 under the equation, similar to addition. repeat but next time when you multiply 2 by 8 add the 1 (from the previous multiplication) to the answer under and put the 1 from 17 on top of the 3. you get it?

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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: