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Doss [256]
3 years ago
11

Use the function below to find f(4). f(x)= 1/3•4^x

Mathematics
1 answer:
kenny6666 [7]3 years ago
5 0

Answer:

f(4) = 85 1/3 or 85.333...

Step-by-step explanation:

To find f(4), you must substitute 4 in place of x whenever x appears:

f(x) = \frac{1}{3} * 4^{x}

f(4) = \frac{1}{3} * 4^{4}

Next, simplify the exponent:

f(4) = \frac{1}{3} * 256

Finally, multiply:

f(4) = 85 1/3 or 85.333...

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Suppose that 35% of all voters voted for a recall. Find the mean of the sampling distribution of the proportion of the 3150 peop
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Answer:

0.35 is the mean of the sampling distribution of the proportion.

Step-by-step explanation:

We are given the following in the question:

Percentage of voters who voted for recall = 35%

p = 35\% = 0.35

Sample size, n = 3150

We have to find the mean of the sampling distribution.

Formula for mean of sampling distribution:

\mu = p

Putting values, we get,

\mu = 0.35

Thus, 0.35 is the mean of the sampling distribution of people sampled in an exit poll who voted for the recall.

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Answer:

I wanna say 1/3 because there 3 even numbers on a dice

Step-by-step explanation:

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Find the surface area of a cylinder with a diameter of 4 cm, and a height of
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125.66

Step-by-step explanation:

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Erin has 5 cookies, and Adlynn has 8. If Mallory has half the number of cookies as Adlynn, how many
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3 years ago
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A company receives shipments of a component used in the manufacture of a component for a high-end acoustic speaker system. When
Tanya [424]

Answer:

ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434

Step-by-step explanation:

For this case we have a sample size of n = 250 units and in this sample they found that 24 units failed one or more of the tests.

We are interested in the proportion of units that fail to meet the company's specifications, and we can estimate this with:

\hat p = \frac{24}{250}= 0.096

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The confidence interval for a proportion is given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.33  

And the margin of error would be:

ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434

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4 years ago
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