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Keith_Richards [23]

3 years ago

7

Riley has collected surveys for her final thesis project. One one question, 46.1% of respondents said that conditions were impro

ving. If Riley calculates the margin of error to be 5.63%, what is the maximum value of the confidence interval for this set of survey results?

1 answer:

ira [324]3 years ago

5 0

Answer:

The maximum value of the confidence interval for this set of survey results is 51.73%.

Step-by-step explanation:

A confidence interval has two bounds, a lower bound and an upper bound.

These bounds depend on the sample proportion and on the margin of error.

The lower bound is the sample proportion subtracted by the margin of error.

The upper bound is the margin of error added to the sample proportion.

In this question:

Sample proportion: 46.1%

Margin of error: 5.63%.

Maximum value is the upper bound:

46.1+5.63 = 51.73

The maximum value of the confidence interval for this set of survey results is 51.73%.

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Is there any options or anything to go with this?

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3 years ago

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

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3 years ago

18 divided by 5593 equals 31-5400 = 193 what number should be placed in the box to help complete the division calculation?

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1 I think should be put in the box.

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3 years ago

Find the least number when divided by 75,125 and 225 Leaves a remainder 6 in each case.

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Answer:

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Step-by-step explanation:

L.C.M of 75,125 and 225,

=>676,125

Leaves a remainder 6 means,

=>676,125 + 6

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That means,

When you divide 676,131 by 75,125 or 225,

You get remainder 6 .

That is why we add 6 to the L.C.M

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3 years ago

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valentinak56 [21]

There are 16 minutes of commercials per hour from 8 A.M to 11 P.M

First, we have to find out how many hours is 8 A.M to 11 P.M

The hours are: 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11.

So, it is 16 hours from 8 A.M to 11 P.M

Then, we multiply multiply 16 by 16 to find out how many minutes of commercials are in 16 hours.

16 x 16 = 256

So, there is 256 minutes of commercials in 16 hours.

Lastly, multiply 256 by 365 to see how many minutes of commercials are in one year.

256 x 365 = 93,440

Therefore, there are 93,440 minutes of commercials in a year.

8 0

3 years ago

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