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dlinn [17]
2 years ago
13

Using a Simple Radom Sample of 500 rental property owners in Washington State, a 95% confidence interval estimate of the proport

ion who own rental property by the water is (0.281, 0.392). Which of the following is the correct interpretation of this interval?
a. Ninety-five percent of all samples of size 500 will result in a proportion of rental property between 0.281 and 0.392.
b. There is probability equal to 0.95 that the proportion of owning a rental property by the water is between 0.281 and 0.392
c. 95% confident that the sample proportion of rental propery owners in Washington is between 28.1% and 39.2%.
d. We can be 95% confident that between 28.1% and 39.2% of all rental property owners own a property by the water
Mathematics
1 answer:
Feliz [49]2 years ago
3 0

Answer:

Hence the correct option is option d.

Explanation:

We can 95% confident that between 28.1% and 39.2% of all rental property owners own a property by the water.  

Here correct option for interpretation of the 95% confidence interval. The remaining options are about probabilities and sample proportions.

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50 plus 30 times 14 equalsss
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Answer:

470 is your awnser

Step-by-step explanation:

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Use the definition of the derviative to compute the derivative of f(x)= 1- 7x^2 at the specific point x=2.
GaryK [48]

Answer:

f'(x) = -28 \ at  \ x = 2

Step-by-step explanation:

f'(2) =  \lim_{h \to 0}\frac{f(2+h)) -f(2)}{h} \\\\f(2+h) = 1 - 7(2+h)^2 = 1 - 7(4 +h^2 +4h) = 1 - 28 - 7h^2 - 28h = - 7h^2 -28h - 27\\\\f(2) = 1 - 7(2^2) = 1 - 28 = -27\\\\f(2+h) - f(2) = -7h^2 - 28h - 27 - (-27) = -7h^2 -28h -27 + 27 = -7h^2 -28h\\\\f'(2) =   \lim_{h \to 0}\frac{-7h^2 - 28h}{h} \\\\

       = \lim_{h \to 0} \frac{h(-7h -28)}{h}\\\\= \lim_{h \to 0} (-7h -28)}\\\\= -7 (0) - 28\\\\= -28

4 0
3 years ago
How many minutes of calling time do the plans cost the same amount ?
Mama L [17]
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7 0
2 years ago
For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[
Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

Solve point 1 that is \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\:

when,

k= 1 \to  s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\

                  = \frac{1}{2} - \frac{1}{3}\\\\

k= 2 \to  s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\

                  = \frac{1}{3} - \frac{1}{4}\\\\

k= 3 \to  s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\

                  = \frac{1}{4} - \frac{1}{5}\\\\

k= n^  \to  s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\

Calculate the sum (S=s_1+s_2+s_3+......+s_n)

S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\

   =\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\

When s_n \ \ dt_{n \to 0}

=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\

\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}

In point 2: \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

when,

k= 1 \to  s_1 = \frac{1}{(1+6)(1+7)}\\\\

                  = \frac{1}{7 \times 8}\\\\= \frac{1}{56}

k= 2 \to  s_1 = \frac{1}{(2+6)(2+7)}\\\\

                  = \frac{1}{8 \times 9}\\\\= \frac{1}{72}

k= 3 \to  s_1 = \frac{1}{(3+6)(3+7)}\\\\

                  = \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\

k= n^  \to  s_n = \frac{1}{(n+6)(n+7)}\\\\

calculate the sum:S= s_1+s_2+s_3+s_n\\

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\

when s_n \ \ dt_{n \to 0}

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066

\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}

8 0
3 years ago
Which exponential functions have been simplified correctly check all that apply
Fittoniya [83]
Principle: Law of Exponents - Combination of product to a power & power to a power. The first is when raising a product of two integers to a power, the power is distributed to each factor. In equation it is,

(xy)^a = (x^a)(y^a)

The latter is when raising the base with a power to a power, the base will remain the same and the powers will be multiplied. In equation it is, 
(x^a)(x^b) = x^ab

Check:


f(x) = 5*(16)^.33x = 5*(8*2)^0.33x = 5*(8^0.33x)(2^0.33x) = 5*(2^x)*(2^0.33x) = 5*(2^1.33x)

f(x) = 2.3*(8^0.5x) = 2.3*(4*2)^0.5x = 2.3*(2^x)(2^0.5x) = 2.3*(2^1.5x)

f(x) = 81^0.25x = 3^x

f(x) = 0.75*(9*3)^0.5x = 0.75*(3^x)*(3^0.5x) = 0.75*3^1.5x

f(x) = 24^0.33x = (8*3)^0.33x = (2^x)*(3^0.33x)


Therefore, the answer is third equation.

<em>ANSWER: f(x) = 81^0.25x = 3^x</em>
5 0
3 years ago
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