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charle [14.2K]
2 years ago
10

The IQ test is designed such that the scores on it are normally distributed with a mean of 100 and a standard deviation of 15. I

f
a person can score two standard deviations above the mean, then they are considered gifted. What score does a person need
in order to be classified as gifted?
Mathematics
1 answer:
kozerog [31]2 years ago
5 0

Answer:

130

Step-by-step explanation:

A standard deviation is 15, so 2 standard deviation is 30, 2 standard deviations above 100 is 100+30=130

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Solve 6y2-5y-6 = 0 using the quadratic formula.
sergejj [24]

Answer:

Ay=3 over 2 or y=-2 over 3

Step-by-step explanation:

the quadratic formula: y={-b±√(b²-4ac)}/2a

In the equation 6y²-5y-6= 0, a=6, b=-5, c= -6

Substituting for the values in the formula we get:

{-(-5)±√[(-5²)-4(6)(-6)}/2(6)

{5±√169}12

={5±13}/12

(5+13)/12=3/2 or (5-13)/12= -2/3

4 0
3 years ago
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How to write 23 minutes after 4
olga_2 [115]
To write 4 as a time you write 4:00,  Then the word after means to add 23 minutes to the 4 so we have 4:23
6 0
3 years ago
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Use PEMDAS to simplify the expression.<br> 7 715-221<br> 2<br> + 10 - (5 x 23)
Bogdan [553]

Answer:

5398

Step-by-step explanation:

7715-2212+10-(5x23)

7715-2212+10-115

5503+10-115

5513-115

5398

3 0
2 years ago
Write the number 3.8 in the forum of a/b using integers to show that it is a rational number
Naya [18.7K]
The answer is b. 38 divided by 10 equals 3.8
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Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos &#10;y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
4 0
3 years ago
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