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Mandarinka [93]
3 years ago
6

I need help I’m confused

Mathematics
1 answer:
Oksana_A [137]3 years ago
5 0

Answer:

11, 18, 25,( 32, 39, 46)

Step-by-step explanation:

The terms are adding up by 7 each so add 7 each following term

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the cost of a school banquet is $95 plus $15 for each person attending. write an equation that gives total cost as a function of
AleksandrR [38]

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8 0
3 years ago
A given population proportion is .25. What is the probability of getting each of the following sample proportions
anyanavicka [17]

This question is incomplete, the complete question is;

A given population proportion is .25. What is the probability of getting each of the following sample proportions

a) n = 110 and = p^ ≤ 0.21, prob = ?

b) n = 33 and p^ > 0.24, prob = ?

Round all z values to 2 decimal places. Round all intermediate calculation and answers to 4 decimal places.)

Answer:

a) the probability of getting the sample proportion is 0.1660

b) the probability of getting the sample proportion is 0.5517

Step-by-step explanation:

Given the data in the questions

a)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 110

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 110 ) √( 0.1875 / 110 ) = 0.0413

Now, P( p^ ≤ 0.21 )

= P[ (( p^ - μ ) /S.D) < (( 0.21 - μ ) / S.D)

= P[ Z < ( 0.21 - 0.25 ) / 0.0413)

= P[ Z < -0.04 / 0.0413]

= P[ Z < -0.97 ]

from z-score table

P( X ≤ 0.21 ) = 0.1660

Therefore, the probability of getting the sample proportion is 0.1660

b)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 33

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 33 ) = √( 0.1875 / 33 ) = 0.0754

Now, P( p^ > 0.24 )  

= P[ (( p^ - μ ) /S.D) > (( 0.24 - μ ) / S.D)

= P[ Z > ( 0.24 - 0.25 ) / 0.0754 )

= P[ Z > -0.01 / 0.0754  ]

= P[ Z > -0.13 ]

= 1 - P[ Z < -0.13 ]

from z-score table

{P[ Z < -0.13 ] = 0.4483}

1 - 0.4483

P( p^ > 0.24 )  = 0.5517

Therefore, the probability of getting the sample proportion is 0.5517

6 0
3 years ago
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