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Pepsi [2]
2 years ago
13

Mr. haigwood is shopping for a school picnic . veggie burgers comes in packages of 15 and buns come in packages of 6. he wants t

o serve veggie burgers on buns and wants to have no items left over. Mr. haighwood says that he will have to buy at least 90 of each item 6 x 15 =90. Do you agree with his reasoning
Mathematics
2 answers:
sveta [45]2 years ago
8 0
No, to find the least number of burgers and buns, you should find out the LCM of 15 and 6, not their product.

LCM = 30,
so they need to buy atleast 30 of each item.
ASHA 777 [7]2 years ago
3 0

Answer:

Mr. haigwood is not right.

Step-by-step explanation:

The veggie burgers comes in packages of 15 and buns come in packages of 6.

Mr. haigwood wants to serve veggie burgers on buns and wants to have no items left over.

Now, to find the least number of burgers and buns needed so nothing is left over, we will find the LCM of 15 and 6.

15 = 3 x 5

6 = 2 x 3

LCM = 2 x 3 x 5 = 30

As given, Mr. haighwood says that he will have to buy at least 90 of each item 6 x 15 =90.

This is not right. He needs to buy at least 30 of each item so that nothing is left over.

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Scores on the SAT Mathematics test are believed to be normally distributed. The scores of a simple random sample of five student
AysviL [449]

Answer:

The mean calculated for this case is \bar X=584

And the 95% confidence interval is given by:

584-2.776\frac{86.776}{\sqrt{5}}=476.271    

584+2.776\frac{86.776}{\sqrt{5}}=691.729    

So on this case the 95% confidence interval would be given by (476.271;691.729)    

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)

s represent the sample 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)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=584

The sample deviation calculated s=86.776

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=5-1=4

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that t_{\alpha/2}=2.776

Now we have everything in order to replace into formula (1):

584-2.776\frac{86.776}{\sqrt{5}}=476.271    

584+2.776\frac{86.776}{\sqrt{5}}=691.729    

So on this case the 95% confidence interval would be given by (476.271;691.729)    

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Step-by-step explanation:

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