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inn [45]
2 years ago
12

71 cards 5 cards in each box 71 ÷5 with leftover How many boxes are needed for all the cards

Mathematics
2 answers:
Allisa [31]2 years ago
8 0
15 boxes are needed because 70÷5=14
with 1 remainder so you will need another box for that one

Fofino [41]2 years ago
7 0

Answer:

15 boxes

Step-by-step explanation:

Because 71 divided by 5 equals to 14.2, you know that you need at least 14 boxes. However, you cannot have 14.2 boxes, because that implies that you have a fifth of a box, which cannot happen. But because each box can only hold up to 5 cards max, you need to get a 15th box for the last card.

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Let the sample size of leg strengths to be 7 and the sample mean and sample standard deviation be 630 watts and 32 watts, respec
Colt1911 [192]

Answer:

a. There is_<u><em>sufficient</em></u> evidence that the leg

C. 0.010 < P-value < 0.025

b. Power of test = 1- β=0.2066

c. So the sample size is 88

Step-by-step explanation:

We formulate the null and alternative hypotheses as

H0 : u1= u2 against Ha : u1 > u2 This is a right tailed test

Here n= 7 and significance level ∝= 0.005

Critical value for a right tailed test with 6 df is 1.9432

Sample Standard deviation = s= 32

Sample size= n= 7

Sample Mean =x`= 630

Degrees of freedom = df = n-1= 7-1= 6

The test statistic used here is

Z = x- x`/ s/√n

Z= 630-600 / 32 / √7

Z= 2.4797= 2.48

P- value = 0.0023890 > ∝ reject the null hypothesis.

so it lies between 0.010 < P-value < 0.025

b) Power of test if true strength is 610 watts.

For  a right tailed test value of z is = ± 1.645

P (type II error) β= P (Z< Z∝-x- x`/ s/√n)

Z = x- x`/ s/√n

Z= 610-630 / 32 / √7

Z=0.826

P (type II error) β= P (Z< 1.645-0.826)

= P (Z> 0.818)

= 0.7933

Power of test = 1- β=0.2066

(c)

true mean = 610

hypothesis mean = 600

standard deviation= 32

power = β=0.9

Z∝= 1.645

Zβ= 1.282

Sample size needed

n=( (Z∝ +Zβ )*s/ SE)²

n=  ((1.645+1.282) 32/ 10)²

Putting the values  and solving we get 87.69

So the sample size is 88

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3 years ago
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3 years ago
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Which graph represents x
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