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erastovalidia [21]

3 years ago

6

The following table shows the percentage of children in the United States between the ages of 3 and 5 who are enrolled in prepri

mary programs.
Date 2008 2010 2012 2014
Percentage 63.0 63.7 64.3 64.7
Use interpolation to estimate the percentage of children enrolled in 2011. (Round your answer to one decimal place.)

1 answer:

docker41 [41]3 years ago

4 0

The percentage of children enrolled in 2011 is 64 %.

<u>Step-by-step explanation:</u>

From the given data we are given with set of x values an y values.

Moreover, the years were given with differences of 2 years.

So let us consider the years 2008 as 0, 2010 as 2, 2012 as 4 and 2014 as 6.

Using interpolation we have to find the percentage of children enrolled in 2011. i.e. 3.

Interpolation is given by the formula,

$\frac{y-y_1}{y_2-y_1} =\frac{x-x_1}{x_2-x_1}$

Let us choose the points (2, 63.7) and (4, 64.3), since 3 is between 2 and 4.

Substitute the values in the formula.

$\frac{y-63.7}{64.3-63.7} =\frac{3-2}{4-2}$ .

$\frac{y-63.7}{0.6} =\frac{1}{2}$ .

$y-63.7 =\frac{1}{2}(0.6).$

y-63.7=0.3.

y= 63.7+0.3

y= 64.

The percentage of children enrolled in 2011 is 64 %.

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A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 sen

tatyana61 [14]

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average desired retirement age = 55 years

$\sigma$ = sample standard deviation = 3.4 years

n = sample of seniors = 101

$\mu$ = true mean retirement age of all college students

<em>Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 96% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.114 < $t_1_0_0$ < 2.114) = 0.96 {As the critical value of t at 100 degree

of freedom are -2.114 & 2.114 with P = 2%}

P(-2.114 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.114) = 0.96

P( $-2.114 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

P( $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

<u>96% confidence interval for</u> $\mu$ = [ $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $55-2.114 \times {\frac{3.4}{\sqrt{101} } }$ , $55+2.114 \times {\frac{3.4}{\sqrt{101} } }$ ]

= [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

7 0

3 years ago

Each piece of candy weighs one eight pound. If Donna buys 15 pieces of candy how many pounds of candy will she buy.

likoan [24]

Answer:

120

Step-by-step explanation:

15/(1/8)

4 0

3 years ago

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n200080 [17]

Answer:

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expression 2: 6m+12+42m

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6(8m+2)= 6(8m+2)

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3 years ago

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Nezavi [6.7K]

Answer:

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Step-by-step explanation:

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A(r) = lw

A(r) = (14)(16)

A(r) = 224 square feet

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A(t) = (8)(14) / 2

A(t) = 112 / 2

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Add the two areas:

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Answer:

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And for this case the value for the height is $h = 10.7 in$ the diameter is provided $D = 2r = 8.1 in$ so then the radius is given by:

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Then we can find the volume with the first formula and replacing we got:

$V = \pi (4.05in)^2 (10.7 in)= 551.4 in ^3$

The final answer for this case would be 551.4 cubic inches

Step-by-step explanation:

The volume for a right circular cone is given by this formula:

$V = \pi r^2 h$

And for this case the value for the height is $h = 10.7 in$ the diameter is provided $D = 2r = 8.1 in$ so then the radius is given by:

$r = \frac{D}{2}=\frac{8.1 in}{2}= 4.05 in$

Then we can find the volume with the first formula and replacing we got:

$V = \pi (4.05in)^2 (10.7 in)= 551.4 in ^3$

The final answer for this case would be 551.4 cubic inches

7 0

3 years ago

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