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Delicious77 [7]
3 years ago
7

The second hand on a clock is

Mathematics
1 answer:
VikaD [51]3 years ago
7 0

Answer:

8.37 cm rounded

Step-by-step explanation:

C = 2π r

C = 2π (8)

C = 16 π

C = 50.24

1/6 ( 50.24)= 8.37

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If 1/4 of x is 16, what is 3/4 of x! Justify your answer in more than one way
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A matched-pairs t-test is NOT an appropriate way to analyze data consisting of which of the following? Measurements of annual in
o-na [289]

Answer:

Step-by-step explanation:

A matched pair t test is one in which the set of data is divided into two sets of data and the test is done to test the difference between the pairs. Sometimes this is called paired t test.\

This is done normally where one group of sample or population can be paired with other group of sample /population.

Hence

A .Measurements of annual income for each twin for 100 randomly selected pairs of twins -- True

B. Measurements of annual income for both individuals in pairs formed by matching 100 people from State A and 100 people from State B based on level of education -- True

C. Measurements of annual income for both individuals in pairs formed by assigning 100 people to pairs at random -- True

D. Measurements of annual income recorded for both spouses of 100 randomly selected married couples  --False here we cannot say paired as these people are not grouped under the same conditions.

7 0
4 years ago
I need help on 11-16
alisha [4.7K]
I cant see all of 11 and 14
12:
3f = 18 \\   \frac{3f}{3}  =  \frac{18}{3}  \\ f = 6
13:
3f =  {3}^{2}  \\ 3f = 9 \\  \frac{3f}{3}  =  \frac{9}{3}  \\ f = 3
15:
3h = 9 + 10 \\ 3h = 19 \\  \frac{3h}{3}  =  \frac{19}{3}  \\ h = 6.3
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6 0
4 years ago
Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find th
alexandr1967 [171]

I'm guessing the function is

f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}

which, split into partial fractions, is equivalent to

\dfrac1{x-6}-\dfrac1{x+2}

Recall that for |x| we have

\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n

With some rearranging, we find

\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n

valid for \left|\dfrac x6\right|, or |x|, and

\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n

valid for \left|-\dfrac x2\right|, or |x|.

So we have

f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n

f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)

f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n

Taken together, the power series for f(x) can only converge for |x|, or -2.

6 0
3 years ago
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