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disa [49]
3 years ago
7

If the mean I.Q. is 100 and the standard deviation of I.Q. scores is 15, then an I.Q. of 130 will have a z score (or standard sc

ore) of ___________________.
Mathematics
1 answer:
skad [1K]3 years ago
7 0

Answer: I.Q. of 130 will have a z score (or standard score) of 2.

Step-by-step explanation:

Let X be a random variable that represents the  I.Q. score.

Formula for Z-score : Z=\dfrac{X-mean}{standard \ deviation}

As per given, mean I.Q. = 100 and the standard deviation =15

For X = 130,

Z=\dfrac{130-100}{15}=\dfrac{30}{15}=2

So, I.Q. of 130 will have a z score (or standard score) of 2.

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8 0
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A cube with side length 5h^2 is stacked on another cube with side length 3k What is the total volume of the cubes in factored fo
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3 years ago
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The volume of a right circular cone with radius r and height h is V = pir^2h/3.
Scorpion4ik [409]

The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

$V=\frac{1}{3} \pi r^2 h$

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9  and dr = 6.8 - 5.9 = 0.9

     h = 4  and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

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So, r = 6.47  and dr = 6.45 - 6.47 = -0.02

     h = 10  and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

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Answer:

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6 0
2 years ago
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