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3 years ago

Choose the product. 4p^3(p^2 +8p-3

1 answer:

5 0

I think you left a bracket which is really important for rhe solution but,these are the possible solutions depending on the bracket.Sorry if I made any mistakes
4p^3(p^2 +8p-3)
=4p^5+32p^4 - 12p^3

(or)
4p^3(p^2) +8p-3
=4p^5+8p-3

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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:

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

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

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

$P(Z < 1.136 ) = 0.87202$

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

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

4 0

2 years ago

What is TU in circle D?<br><br> 5.1<br> 3.4<br> ا<br> O 3.4<br> O 3.8<br> O 7.6<br> O 5.1

Debora [2.8K]

Answer:

answer-3.4

Step-by-step explanation:

2022 edge

8 0

2 years ago

How does the Binomial Theorem’s use Pascal’s triangle to expand binomials raised to positive integer powers?

MrMuchimi

Answer:

There are ways for quickly multiply out a binomial that's being raised by an exponent. Like

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = (a + b)(a + b)2 = (a + b)(a2 + 2ab + b2) = a3 + 3a2b + 3ab2 + b3

and so on and so on

but there was this mathematician named Blaise Pascal and he found a numerical pattern, called Pascal's Triangle, for quickly expanding a binomial like the ones from earlier. It looks like this

1 1

2 1 2 1

3 1 3 3 1

4 1 4 6 4 1

5 1 5 10 10 5 1

Pascal's Triangle gives us the coefficients for an expanded binomial of the form (a + b)n, where n is the row of the triangle.

Hope this helps!

5 0

2 years ago

The curved part of this figure is a semicircle.

ikadub [295]

Answer:

The actual answer is 10.5+7.25π units²

I just took the test

5 0

2 years ago

A regular six-sided die is rolled 1000 times. Use the binomial distribution to determine the standard deviation for the number o

ANTONII [103]

The standard deviation for the number of times an odd number is rolled is 15.8

<h3>How to determine the standard deviation?</h3>

The given parameters are:

Die = regular six-sided die

n = 1000

The probability of rolling an odd number is:

p = 1/2 = 0.5

The standard deviation is then calculated as;

$\sigma = \sqrt{np(1 - p)$

This gives

$\sigma = \sqrt{1000 * 0.5 * (1 - 0.5)$

Evaluate the products

$\sigma = \sqrt{250$

Evaluate the root

$\sigma = 15.8$

Hence, the standard deviation is 15.8

Read more about standard deviation at:

brainly.com/question/16555520

#SPJ1

8 0

1 year ago