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Pepsi [2]
3 years ago
8

Mariam’s uncle donate 120 cans of juice and 90 packs of cheese cracker for the school picnic. Each student must receive the same

number of cans of juice and the same numbers of pack of cracker with no leftover. Suppose Mariam Uncle eats two pack of the cracker before he brings the supplies to school. What is the greatest number of student who can share the food equally? How many cans of juice and packs of cracker will each student receive? Explain
Mathematics
1 answer:
monitta3 years ago
4 0

Answer:

8 students,

15 cans of juice and 11 packs of cheese cracker

Step-by-step explanation:

Mariam’s uncle donate 120 cans of juice and 90 - 2 = 88 packs of cheese cracker (eats two packs) for the school picnic.

Factor these numbers:

120=2\cdot 60=2\cdot 2\cdot 30=2\cdot 2\cdot 2\cdot15=2\cdot 2\cdot 2\cdot 3\cdot 5\\ \\88=2\cdot 44=2\cdot 2\cdot 22=2\cdot 2\cdot 2\cdot 11

The greatest common factor is

2\cdot 2\cdot 2=8,

so the greatest number of students who can share the food equally is 8 and each student will get 3\cdot 5=15 cans of juice and 11 packs of cheese cracker.

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Answer with explanation:

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The largest degree of numerator is 7 , while the largest degree of denominator is 4.So, as we know

 \frac{x^a}{x^b}=x^{a-b}\\\\\frac{x^7}{x^4}=x^{7-4}\\\\=x^3

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8 0
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Answer:

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

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c) 95% of confidence intervals are  (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes n_{1} = 400 and n_{2} = 300

Proportion of mean p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52

<u>Null hypothesis H0</u> : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

<u>Alternative hypothesis H1:</u>- p1 ≠ p2

a) The test statistic is

Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }

where p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}  

on calculation we get   p = 0.56    

now q =1-p = 1-0.56=0.44

Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }\\   =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300}   }

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

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p1 = p2

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The confidence intervals are P± 1.96(√PQ/n)

we know that = p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}

after calculation we get P = 0.56 and Q =1-P =0.44

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now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of <u>confidence intervals  (0.523 ,0.596)</u>

<u></u>

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Answer:

that is the solution

Step-by-step explanation:

it cannot be simplified

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