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qwelly [4]
3 years ago
8

Find the sum to infinity​

Mathematics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

-7/27

Step-by-step explanation:

-14/54

Hope this helped :)

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The number of violent crimes committed in a day possesses a distribution with a mean of 2.8 crimes per day and a standard deviat
Deffense [45]

Answer:

The distribution is approximately normal with mean = 2.8 and standard error = 0.4

Step-by-step explanation:

We are given;

Mean; μ = 2.8

Standard deviation; σ = 4

Sample size; n = 100

Now, the central limit theorem states that the sample mean with a sample size(n) from a population mean (μ) and population standard deviation(σ) will, for large value of n, have an approximately normal distribution with mean μ and standard error as (σ/√n)

The sample size is 100 and thus it's very large because it's bigger than minimum of 30 for approximate distribution.

Thus, SE = (σ/√n) = 4/√100 = 4/10 = 0.4

Thus,from the central limit theorem I described, we can say that the distribution is approximately normal with mean = 2.8 and standard error = 0.4

8 0
3 years ago
A bus eats 4 people in a row and there are 12 rows how would fit on 2 buses?
BARSIC [14]

Answer:

4 \times 12 \times 2 = 96

96 people in total.

BTW its seats and I think you meant "how many"

6 0
3 years ago
Why is it important to avoid a smaller sample size?
Nitella [24]

Answer:The importance of sample size calculation cannot be overemphasized. A research can be conducted for various objectives. A smaller sample will give a result which may not be sufficiently powered to detect a difference between the groups and the study may turn out to be falsely negative leading to a type II error.

5 0
3 years ago
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The mad scientist will need two fifth of the potions how many potion is that
gulaghasi [49]
40% if you are asking for percent that’s you answer
7 0
3 years ago
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What is the value of this expression when a = 3 and b = -2?
weqwewe [10]

The first step: Simplify the expression with:

\dfrac{x^n}{x^m}=x^{n-m}\\\\(xy)^n=x^ny^n\\\\(x^n)^m=x^{nm}\\\\x^{-n}=\dfrac{1}{x^n}

\left(\dfrac{3a^{-2}b^6}{2a^{-1}b^5}\right)^2=\left(\dfrac{3}{2}a^{-2-(-1)}b^{6-5}\right)^2=\left(\dfrac{3}{2}a^{-1}b^1\right)^2\\\\=\left(\dfrac{3}{2}\right)^2(a^{-1})^2b^2=\dfrac{9}{4}a^{-2}b^2=\dfrac{9b^2}{4a^2}

The second step: Put the values of a = 3 and b = -2 to the expression:

\dfrac{9(-2)^2}{4(3)^2}=\dfrac{(9)(4)}{(4)(9)}=1

<h2>Answer: 1</h2>
4 0
4 years ago
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