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3 years ago

Find x. X to the power of 3 = 1/8

Mathematics

1 answer:

xenn [34]3 years ago

6 0

Mabye 1/2 i think it is that let me check

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You recently sent out a survey to determine if the percentage of adults who use social media has changed from 66%, which was the

il63 [147K]

Answer:

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of the 2809 people who responded to survey, 1634 stated that they currently use social media.

This means that $n = 2809, \pi = \frac{1634}{2809} = 0.5817$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 - 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.56$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 + 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.6034$

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

6 0

3 years ago

Fifty-three percent of employees make judgements about their co-workers based on the cleanliness of their desk. You randomly sel

GarryVolchara [31]

Answer:

The unusual $X$ values for this model are: $X = 0, 1, 2, 7, 8$

Step-by-step explanation:

A binomial random variable $X$ represents the number of successes obtained in a repetition of $n$ Bernoulli-type trials with probability of success $p$ . In this particular case, $n = 8$ , and $p = 0.53$ , therefore, the model is ${8 \choose x} (0.53) ^ {x} (0.47)^{(8-x)}$ . So, you have:

$P (X = 0) = {8 \choose 0} (0.53) ^ {0} (0.47) ^ {8} = 0.0024$

$P (X = 1) = {8 \choose 1} (0.53) ^ {1} (0.47) ^ {7} = 0.0215$

$P (X = 2) = {8 \choose 2} (0.53)^2 (0.47)^6 = 0.0848$

$P (X = 3) = {8 \choose 3} (0.53) ^ {3} (0.47)^5 = 0.1912$

$P (X = 4) = {8 \choose 4} (0.53) ^ {4} (0.47)^4} = 0.2695$

$P (X = 5) = {8 \choose 5} (0.53) ^ {5} (0.47)^3 = 0.2431$

$P (X = 6) = {8 \choose 6} (0.53) ^ {6} (0.47)^2 = 0.1371$

$P (X = 7) = {8 \choose 7} (0.53) ^ {7} (0.47)^ {1} = 0.0442$

$P (X = 8) = {8 \choose 8} (0.53)^{8} (0.47)^{0} = 0.0062$

The unusual $X$ values for this model are: $X = 0, 1, 7, 8$

6 0

4 years ago

Read 2 more answers

Four ninths plus one ninth?

Kazeer [188]

Answer:

5/9

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=5%28sin%28t%29%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%2Bycos%28t%29%29%29%3Dcos%28t%29%20%20%28sin%28

Nataliya [291]

Assuming you mean

$5\sin t\dfrac{\mathrm dy}{\mathrm dt}+5y\cos t=\cos t\sin^2t$

This ODE is linear in $y$ , and you can already contract the left hand side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[5y\sin t\right]=\cos t\sin^2t$

Integrating both sides with respect to $t$ yields

$5y\sin t=\displaystyle\int\cos t\sin^2t\,\mathrm dt$
$5y\sin t=\dfrac13\sin^3t+C$
$y=\dfrac1{15}\sin^2t+C\csc t$

Given that $y\left(\dfrac\pi2\right)=9$ , we have

$9=\dfrac1{15}\sin^2\dfrac\pi2+C\csc\dfrac\pi2$
$9=\dfrac1{15}+C$
$C=\dfrac{134}{15}$

so that the particular solution over the interval is

$y=\dfrac1{15}\sin^2t+\dfrac{134}{15}\csc t$

6 0

3 years ago

Sarah took the advertising department from her company on a round trip to meet with a potential client. Including Sarah a total

ololo11 [35]

Answer:

Step-by-step explanation:

c + f = 16......c = 16 - f

340c + 910f = 9430

340(16 - f) + 910f = 9430

5440 - 340f + 910f = 9430

-340f + 910f = 9430 - 5440

570f = 3990

f = 3990/570

f = 7 <====== 7 first class tickets bought

c + f = 16

c + 7 = 16

c = 16 - 7

c = 9 <====== 9 coach tickets bought

3 0

3 years ago