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lina2011 [118]
2 years ago
7

The average MCAT score follows a Normal distribution, with a mean of μ = 508 and a standard deviation of σ = 8. What is the prob

ability that the mean IQ score of 100 randomly selected people will be more than 510?
0.0062
0.9938
0
0.4013
0.5987
Mathematics
2 answers:
dusya [7]2 years ago
8 0

Subtract the mean from the high score and divide by the standard deviation:

510 - 508 = 2

2/8 = 0.25

Using the Z tabe find the z value for 0.25 = 0.5987

Now subtract that value from 1:

1 - 0.5987 = 0.4013

tatyana61 [14]2 years ago
3 0

Answer:

0.4013

Step-by-step explanation:

(Just want to note I am not the best at explanations but here is what I came up with to save you some time)

I watched a vid with sample questions so i realized I needed to get my graphing calculator and input my info into normal cdf

<u>Inputing</u><u> </u>

(510,999,508,8) = 0.40129....... = <u><em>.4013</em></u>

Hope this helped :)

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martin will roll a number of cube 150 times.The number cube is labeled 1 to 6.How many times could Martin expect to roll a numbe
olga_2 [115]
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2 years ago
What is the volume of the prism? 1080 cm 2700 cm 3240 cm 6480 cm
makkiz [27]

Answer:

Volume of prism = 3,240 cm³

Step-by-step explanation:

GIven.

Hexagonal prism.

Side of base(b) = 12cm

Height of prism = 9cm

Height of base (h)= 10cm

Find:

The volume of the prism.

Computation:

Area of base of hexagonal prism = n/2[bh]

Area of base of hexagonal prism = 6/2[(12)(10)]

Area of base of hexagonal prism = 360 cm²

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3 0
2 years ago
A flat circular plate has the shape of the region x squared plus y squared less than or equals 1x2+y2≤1. the​ plate, including t
vredina [299]

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The target function has partial derivatives (set equal to 0)

\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2

\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0

so there is only one critical point at \left(-\dfrac{13}2,0\right). But this point does not fall in the region x^2+y^2\le1. There are no extreme values in the region of interest, so we check the boundary.

Parameterize the boundary of x^2+y^2\le1 by

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y=\sin u

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t(x,y)=t(\cos u,\sin u)=T(u)

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T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0

(1)\quad\sin u=0\implies u=0,u=\pi

(2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4

but |\cos u|\le1 for all u, so this case yields nothing important.

At these critical points, we have temperatures of

T(0)=14

T(\pi)=-12

so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.

4 0
2 years ago
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