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Rom4ik [11]
3 years ago
10

A newsletter publisher believes that above 41A% of their readers own a personal computer. Is there sufficient evidence at the 0.

100.10 level to substantiate the publisher's claim?
Mathematics
1 answer:
klemol [59]3 years ago
6 0

Complete Question

A newsletter publisher believes that above 41% of their readers own a personal computer. Is there sufficient evidence at the 0.10 level to substantiate the publisher's claim?

State the null and alternative hypotheses for the above scenario.

Answer:

The null hypothesis is  H_o  : p \le   0.411

The  alternative hypothesis  is  H_a : \mu >  0.41

Step-by-step explanation:

From the question we are told that

   The  proportion is  p  =  0.41

Generally the null hypothesis is  H_o  : p \le   0.411

The  above condition represents the  null hypothesis because it contains and equality in its condition

  The  alternative hypothesis  is  H_a : \mu >  0.41

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The short sides of a rectangle are 2 inches. The long sides of the same rectangle are three less than an unknown number of inche
Amiraneli [1.4K]

<u>Answer</u>

The first student was right.

The length of the long side is at least 13 inches.

<u>Explanation</u>

The perimeter of any figure is the distance all round.

Perimeter of a rectangle = 2(l+w). Where l is length and w is the width.

2(l + w) ≥ 30

2{(x-3) + 2} ≥ 30

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x - 1 ≥ 15

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The length of the long side is 13 inches. The first student was right.

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3 years ago
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Solve for n.<br> n+ 1 = 4(n-8)<br> On=1<br> O n= 8<br> on = 11<br> On = 16
Lesechka [4]

Answer:

n = 11

Step-by-step explanation:

n+ 1 = 4(n-8)

<=> n + 1 = 4n - 4×8

<=> 4n - n = 1 + 4×8

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3 0
3 years ago
If you can purchase 5 pounds for $4, how many pounds can you purchase for $16? *
Julli [10]

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8 0
3 years ago
7 freshmen, 9 sophomores, 8 juniors, and 8 seniors are eligible to be on a committee.
Setler79 [48]

Answer:

i)32C16

ii)1185408

<em><u>Explanation</u></em><em><u>:</u></em>

i)Total number of selected/eligible is 7+9+8+8=32

Total ways of selecting dance committee of 16 is

<em><u>3</u></em><em><u>2</u></em><em><u>C</u></em><em><u>1</u></em><em><u>6</u></em>

ii)Total ways of selecting 3 seniors from 8 is 8C3

and Total ways of selecting 6 juniors from 8 is 8C6

ways of selecting 2 sopho from 9 is 9C2

ways of selecting 5 freshman from 7 is 7C5

now, total way of selection come to be

8C3×8C6×9C2×7C5

=56×28×36×21

=1185408

✌️

6 0
3 years ago
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