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3 years ago

How do I solve -1/3y-6=-11

Mathematics

2 answers:

gtnhenbr [62]3 years ago

3 0

Answer:

Step-by-step explanation:

-1/3y - 6 = -11

-1/3y - 6 + 6 = -11 + 6

-1/3y = -5

-1/3y * 3y = -5 * 3y

-1 = -15y

-1 / -15 = -15y / -15y

y = 1/15

Oxana [17]3 years ago

3 0

For this case we must solve the following equation:

$- \frac {1} {3}y-6 = -11$

We add 6 to both sides of the equation:

$- \frac {1} {3} y = -11 + 6\\- \frac {1} {3} y = -5$

We multiply by 3 on both sides of the equation:

$-y = 3 * (- 5)\\-y = -15$

Finally we have:

$y = 15$

Answer:

$y = 15$

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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

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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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A recent study of 26 city residents showed that the time they had lived at their present address has a mean of 10.3 years with t

svetlana [45]

Answer:

The 95% confidence interval is $9.15< \mu < 11.45$

Step-by-step explanation:

From the question we are told that

The sample size is n = 26

The mean is $\= x = 10.3$

The standard deviation is $s = 3$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * \frac{3}{\sqrt{26} }$

=> $E = 1.1532$

Generally 95% confidence interval is mathematically represented as

$\= x -E < \mu < \=x +E$

=> $10.3 -1.1532 < \mu < 10.3 + 1.1532$

=> $9.15< \mu < 11.45$

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Answer:

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Step-by-step explanation:

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