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3 years ago

7

Mr. Monroe keeps a bag of small prizes to distribute to his students. He likes to keep at least twice as many prizes in the bag

as he has students. The bag currently has 79 prizes in it. Mr. Monroe has 117 students. How many more prizes does he need to buy?

1 answer:

Elden [556K]3 years ago

6 0

Answer:

I think the answer is (Mr. Monroe needs to buy 76 more)

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Select all the rational numbers

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Answer:

B. 3.14

C. .33

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3 years ago

The ratio of Sam's money to Buffy's is 1.2. If Sam has $3, how much money does Buffy have? Show your work please

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3 years ago

One environmental group did a study of recycling habits in a California community. It found that 75% of the aluminum cans sold i

AysviL [449]

Answer:

a

$P( \^ p > 0.775 ) = 0.12798$

b

$P( 0.6718 < p < 0.775 ) =0.87183$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.75$

Considering question a

The sample size is $n = 387$

Generally the standard deviation of this sampling distribution is

$\sigma = \sqrt{ \frac{p(1 - p)}{ n } }$

=> $\sigma = \sqrt{ \frac{0.75(1 - 0.75)}{ 387 } }$

=> $\sigma = 0.022$

The sample proportion of cans that are recycled is

$\^ p = \frac{ 300}{387 }$

=> $\^ p = 0.775$

Generally the probability that 300 or more will be recycled is mathematically represented as

$P( \^ p > 0.775 ) = P( \frac{\^ p - p }{ \sigma } > \frac{0.775 - 0.75 }{ 0.022} )$

$\frac{\^ p - p }{\sigma } = Z (The \ standardized \ value\ of \ \^ p )$

$P( \^ p > 0.775 ) = P( Z > 1.136 )$

From the z table the area under the normal curve to the left corresponding to 1.591 is

$P( Z > 1.136) = 0.12798$

=> $P( \^ p > 0.775 ) = 0.12798$

Considering question b

Generally the lower limit of sample proportion of cans that are recycled is

$\^ p_1 = \frac{ 260 }{387 }$

=> $\^ p_1 = 0.6718$

Generally the upper limit of sample proportion of cans that are recycled is

$\^ p_2 = \frac{ 300}{387 }$

=> $\^ p_2 = 0.775$

Generally probability that between 260 and 300 will be recycled is mathematically represented as

$P( 0.6718 < p < 0.775 ) = P( \frac{0.6718 - 0.75 }{ 0.022}< \frac{\^ p - p }{ \sigma }$

=> $P( 0.6718 < p < 0.775 ) = P( -3.55 < Z < 1.136 )$

=> $P( 0.6718 < p < 0.775 ) = P(Z < 1.136 ) - P( Z < -3.55 )$

From the z table the area under the normal curve to the left corresponding to 1.136 and -3.55 is

$P( Z < -3.55 ) = 0.00019262$

and

$P(Z < 1.136 ) = 0.87202$

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$P( 0.6718 < p < 0.775 ) = 0.87202- 0.00019262$

=> $P( 0.6718 < p < 0.775 ) =0.87183$

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NISA [10]

Plug in 0
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