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aksik [14]
3 years ago
6

Suppose there is a pile of quarters dimes and pennies with a total value of $1.07 how much of each coin can be present without b

eing able to make change for a dollar? Of there are multiple selections of the coins that will work choose the selection with the largest total numbers of coins.
Mathematics
1 answer:
Korolek [52]3 years ago
4 0
Hello,

Very nice as problem.

2 solutions:
1 quater,8 dimes, 2 pennies
and
3 quaters,3 dimes, 2 pennies

since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0



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The front and back covers of a text book are each 0.3 thick between the cover are 200 sheets of paper each 0.008 cm thick how hi
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Yolanda spends 8/23 hours per month playing soccer. Approximately how many hours does she play soccer in a year
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Given:
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A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If
musickatia [10]

Answer:

a. The test statistic for this hypothesis would be z=1.71

b. Critical value at α = 0.10: z=1.282

c. Reject H0, the television network should not keep its current lineup.

d. H0: p ≤ 0.50; HA: p > 0.50

Step-by-step explanation:

We have to performa an hypothesis test on a proportion.

The claim is that the proportion of viewers that prefer the new show is bigger than 50% (meaning that most viewers prefer the new show).

The null hypothesis will state that both shows have the same proportion.

Then, the null and alternative hypothesis are:

H_0: \pi\leq0.50\\\\H_a: \pi>0.5

The sample proportion is p=0.53

p=X/n=438/827=0.53

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.5*0.5}{827}}= 0.017

Then, the z-statistic can be calculated as:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.53-0.50-0.5/827}{0.017}=\dfrac{ 0.029 }{0.017} =1.706

The critical value for a right tailed test at a significance level of 0.10 is zc=1.282. This value can be looked up in a standard normal distribution table.

As the statistic is bigger than the critical value, it lies in the rejection region. The null hypothesis is rejected. There is evidence to support the claim that the proportion of viewers which support the new show is larger than 0.50.

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