^ 3 sqrt 750 + ^ 3 sqrt 2058 - ^ 3 sqrt 48
Rewriting the expression we have
^ 3 sqrt (6 * x ^ 3) + ^ 3 sqrt (6 * y ^ 3) - ^ 3 sqrt (6 * z ^ 3)
That is, we have the following equations:
6 * x ^ 3 = 750
6 * y ^ 3 = 2058
6 * z ^ 3 = 48
Clearing x, y and z we have:
x = 5
y = 7
z = 2
Then, rewriting the expression
x (^ 3 sqrt (6)) + y (^ 3 sqrt (6)) - z (^ 3 sqrt (6))
Substituting the values
5 (^ 3 sqrt (6)) + 7 (^ 3 sqrt (6 *)) - 2 (^ 3 sqrt (6))
10 (^ 3 sqrt (6))
answer
the simple form of the expression is
D) 10 ^ 3 sqrt 6
A sector is the area bounded by two radii and the included arc.
To calculate for the area of the sector, you’ll use the formula A=n/360(pi)(radius)

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The n in the formula refers to the measure of the central angle in degrees.
The given data in the problem is in radians, convert it by multiplying 1.25 by
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.
Now input the converted value to the formula of finding the area of the sector.<span>
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Since x represents the length (in meters), the width, which is 7 meters greater, is x +7. The product of these dimensions is the area (in square meters):
x(x +7) = 170
In standard form, the quadratic is ...
x² +7x -170 = 0
Answer:
The probability is 1/2
Step-by-step explanation:
The time a person is given corresponds to a uniform distribution with values between 0 and 100. The mean of this distribution is 0+100/2 = 50 and the variance is (100-0)²/12 = 833.3.
When we take 100 players we are taking 100 independent samples from this same random variable. The mean sample, lets call it X, has equal mean but the variance is equal to the variance divided by the length of the sample, hence it is 833.3/100 = 8.333.
As a consecuence of the Central Limit Theorem, the mean sample (taken from independant identically distributed random variables) has distribution Normal with parameters μ = 50, σ= 8.333. We take the standarization of X, calling it W, whose distribution is Normal Standard, in other words

The values of the cummulative distribution of the Standard Normal distribution, lets denote it
, are tabulated and they can be found in the attached file, We want to know when X is above 50, we can solve that by using the standarization
