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sammy [17]
2 years ago
13

In a survey of first graders, their mean height was 51.6 inches with a standard deviation of 3.6 inches. Assuming the heights ar

e normally distributed, what height represents the first quartile of these students?
Mathematics
1 answer:
ki77a [65]2 years ago
6 0

Answer:

Step-by-step explanation:

In a survey of first graders, their mean height was 51.6 inches with a standard deviation of 3.6 inches. Assuming the heights are normally distributed, what height represents the first quartile of these students?

54.03 inches

48.57 inches

48.00 inches

49.17 inches

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I need help please and thank you
Vaselesa [24]

Answer:

Eagles

Step-by-step explanation:

2000x6=12000 this is the most

5 0
3 years ago
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five states in a country have 4 star hotels. Assume the country has 50 states. (a) what fraction of the states can claim at leas
mojhsa [17]

Part A:

Since five states have four-star hotels, only those five states can claim at least one four-star hotel and since there are 50 states, we can write this as 5/50 or 1/10.

Part B:

50 - 5 = 45 states don't have a four-star hotel.

Part C:

45/50 or 9/10 of the states do not have a four-star hotel.

Best of Luck!

8 0
3 years ago
These are the first four terms of a sequence.
DedPeter [7]

Answer:

2+3n

Step-by-step explanation:

This question is based on a topic called sequence and series,under which we have the arithmetic progression

The formula for arithmetic progression is a+(n-1)d

In which a is the first number,n is the nth term and d is the difference

Back to the question

a=5,n=it's not given,d=3

Using the general formula

a+(n-1)d

5+(n-1)3

Open the bracket first

5+3n-3

Collect like terms

5-3+3n

2+3n

The answer is 2+3n

And it can be solved further only if we are given the value of n...the question even stated that we should solve in terms of n

Therefore the answer is 2+3n

3 0
3 years ago
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Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

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2 years ago
If h(x)=6-x what is the value of (hxh)(10)?
Ghella [55]

Answer:

16

Step-by-step explanation:

h(x) × h(x) = (6 - x)²

(h × h)(10) = (6 - 10)² = (- 4)² = 16

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