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Alenkasestr [34]
2 years ago
11

Use pic for more info

Mathematics
1 answer:
padilas [110]2 years ago
7 0

Answer: The answer is D

Step-by-step explanation: Hope that this helps.

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About 5% of the population has a particular genetic mutation. 300 people are randomly selected.
xz_007 [3.2K]

The standard deviation for the number of people with the genetic mutation is 3.77

<h3>How to determine the standard deviation?</h3>

The given parameters are:

Sample size, n = 300

Proportion that has the particular genetic mutation, p = 5%

The standard deviation for the number of people with the genetic mutation is calculated as:

Standard deviation = √np(1 - p)

Substitute the known values in the above equation

Standard deviation = √300 * 5% * (1 - 5%)

Evaluate the product

Standard deviation = √14.25

Evaluate the exponent

Standard deviation = 3.77

Hence, the standard deviation for the number of people with the genetic mutation is 3.77

Read more about standard deviation at

brainly.com/question/12402189

#SPJ1

7 0
1 year ago
Simplidy y square root 6 over 6 square root y​
Contact [7]

Answer:

Step-by-step explanation:

You can multiply the fraction by 6 and then you are able to cancel out and find the simplification

4 0
2 years ago
Please help me<br><br> b minus 3/4 = 7/10
RoseWind [281]
B - 3/4 = 7/10
add 3/4 to both sides to isolate the b
b = 7/10 + 3/4
b = 28/40 + 30/40
b = 58/40
b = 1  18/40
b = 1   9/20
5 0
3 years ago
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Is 2 hours and 30 minutes more or less than 10% of a day?
masha68 [24]
More then 10% of the day
8 0
3 years ago
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PLZ HELP!!! Use limits to evaluate the integral.
Marrrta [24]

Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

r_i=\dfrac{2i}n

where 1\le i\le n. Each interval has length \Delta x_i=\frac{2-0}n=\frac2n.

At these sampling points, the function takes on values of

f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}

We approximate the integral with the Riemann sum:

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3

Recall that

\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4

so that the sum reduces to

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}

Just to check:

\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28

4 0
2 years ago
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