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2 years ago

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Combining like terms

1 answer:

MAXImum [283]2 years ago

4 0

Combine like terms to simplify an expression. For example, all terms with the variable x can be combined into one term. All constants can also be combined.

1) -4x - 10x = -14x

2) -r - 10r = -11r

3) -2x + 11 +6x = 4x + 11

4) 11r - 12r = -r

5) -v + 12v = 11v

6) -8x - 11x = -19x

7) 4p + 2p = 6p

8) 5n + 11n = 16n

9) n + 4 - 9 - 5n = -4n - 5

10) 12r + 5 + 3r - 5 = 15r (the 5 and -5 cancel each other out)

11) -5 + 9n + 6 = 9n + 1

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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

$P(Z < 1.136 ) = 0.87202$

So

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

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

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