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velikii [3]
3 years ago
5

Scores on an exam are normally distributed with a mean of 76 and a standard deviation of 10. In a group of 230 tests, how many s

tudents score above 96?
A. about 31 students
B.about 14 students C.about 6 students D.about 2 students
Mathematics
1 answer:
sergeinik [125]3 years ago
5 0
C) about 6 students, This is because each standard deviation is 10 so two deviations up = 96, meaning that 2.3% of the people scored above and 2.3% times 230 people is 5.29, so about 6 students.
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Question 12 (2 points)
elena55 [62]

Answer:

A box plot is drawn with end points at 24 and 49.The box extends from 28 to 44 and a vertical line is drawn inside the box at 34.

Step-by-step explanation:

Ordering the data given :

24,28,32,34,40,44,49

We can calculate the 5 number summary required to give the appropriate boxplot that can be produced :

Minimum = 24

Maximum = 49

Median = 1/2(n+1)th term

n = 7

Median = 1/2(8) = 4th term

Median = 34

Lower quartile, Q1 = 1/4(n+1)th term

n = 7

1/4(8) = 2nd term

Q1 = 28

Upper quartile : 3/4(n+1)th term

n = 7

Q3 = 3/4(8) = 6th term

Q3= 44

3 0
3 years ago
I am 21 years old my father is 3 times my age how old eill i be when my father is 40
DENIUS [597]

Answer:

Well the answer is 13.3 so 13 years old

Step-by-step explanation:

(F,Y) = (63,21)

Y3=F so F/3=y

so

F= 40

Y=?

40/3=Y

40/3=13.3

4 0
2 years ago
Read 2 more answers
When y = 35,x = 2.5. If the value of y varies directly with x what is the value of y when the value of xis 3.25
leonid [27]

Answer:

The value of y would be 45.5

Step-by-step explanation:

To solve this problem, start with the base form of direct variation.

y = kx

Now we can use our original values to model the equation and find k.

35 = k(2.5)

14 = k

Now we can model the equation as:

y = 14x

Now to find y, when x = 3.25, simply put 3.25 into the equation.

y = 14(3.25)

y = 45.5

8 0
3 years ago
Find the volume of the solid.
dmitriy555 [2]

In Cartesian coordinates, the region (call it R) is the set

R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}

In the plane z=2, we have

2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2

which is a circle with radius \sqrt2. Then we can better describe the solid by

R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}

so that the volume is

\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx

While doable, it's easier to compute the volume in cylindrical coordinates.

\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}

Then we can describe R in cylindrical coordinates by

R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}

so that the volume is

\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}

3 0
1 year ago
John gets paid $6 for each of the first 40 toy cars he makes in a week. For any addition toy cars beyond 40, his pay increases b
jeka57 [31]
48 times 6 and that will be your answer
3 0
3 years ago
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