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DochEvi [55]
3 years ago
15

Need help again .........

Mathematics
1 answer:
xenn [34]3 years ago
3 0

Answer:

(:

Step-by-step explanation:

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In which branch of statistics would a researcher acquire twenty-five 2000 Toyota Celicas, drive them until they had a major mech
AURORKA [14]

Answer:

The answer is "descriptive statistics".

Step-by-step explanation:

Descriptive statistics are short descriptive factors that add up a production set, that can be either a reflection of a whole or a sample population. It is broken down into cumulative frequency measurements and variance measurements. The study uses descriptive statistics to explain the profile of the respondents. People give simple summaries about the sample as well as the actions.

6 0
3 years ago
Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference
Bezzdna [24]

Since a_1,a_2,a_3,\cdots are in arithmetic progression,

a_2 = a_1 + 2

a_3 = a_2 + 2 = a_1 + 2\cdot2

a_4 = a_3+2 = a_1+3\cdot2

\cdots \implies a_n = a_1 + 2(n-1)

and since b_1,b_2,b_3,\cdots are in geometric progression,

b_2 = 2b_1

b_3=2b_2 = 2^2 b_1

b_4=2b_3=2^3b_1

\cdots\implies b_n=2^{n-1}b_1

Recall that

\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n

\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2

It follows that

a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n +  n(n-1)

so the left side is

2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n

Also recall that

\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}

so that the right side is

b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)

Solve for c.

2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}

Now, the numerator increases more slowly than the denominator, since

\dfrac{d}{dn}(2n(n-1)) = 4n - 2

\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2

and for n\ge5,

2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2

This means we only need to check if the claim is true for any n\in\{1,2,3,4\}.

n=1 doesn't work, since that makes c=0.

If n=2, then

c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0

If n=3, then

c = \dfrac{12}{2^3 - 6 - 1} = 12

If n=4, then

c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N

There is only one value for which the claim is true, c=12.

3 0
2 years ago
Factor completely<br> x^2 - 8x + 16
Anarel [89]

Answer:

(x - 4)^2

Step-by-step explanation:

6 0
3 years ago
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Wdym by animal testing? Like doing experiments on rats as an example?
4 0
3 years ago
NEED HELP ASAP!!!!!!!
Elan Coil [88]
I think the answer is C
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