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2 years ago

(50) POINTS AND 5 STAR AND THANKS AND BRAIN-LIST

Mathematics

1 answer:

Elenna [48]2 years ago

6 0

Answer:

B-15 pounds

Step-by-step explanation:

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Based on historical data, your manager believes that 26% of the company's orders come from first-time customers. A random sample

scoundrel [369]

Answer:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

Step-by-step explanation:

For this case we have the following info given:

$p = 0.26$ represent the proportion of the company's orders come from first-time customers

$n=158$ represent the sample size

And we want to find the following probability:

$p(\hat p >0.4)$

And we can use the normal approximation since we have the following two conditions:

1) np = 158*0.26 = 41.08>10

2) n(1-p) = 158*(1-0.26) = 116.92>10

And for this case the distribution for the sample proportion is given by:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

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Mary has g+24.50 in her bank account. bobby has 7g- 52.34 in his bank account. How much do they have in all?

vova2212 [387]

Combine:

<span>g+24.50

</span><span>7g- 52.34
---------------
8g - 27.84 <= answer</span>

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Answer:

Step-by-step explanation:

$Q+p+Q+p+Q\\\\\text{combine like terms}\\\\=(Q+Q+Q)+(p+p)=3Q+2p$

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