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3 years ago

Solve for x: 3|x-3| 2=14 a. No solution b. x=-1,x=8.3 c. x=0,x=7 d. x=-1,x=7

1 answer:

3 0

I am looking on the answers, and there is only one case, when a or b or c or d pass: 3|x-3| + 2 = 14. So I assume, that before two is plus. Then:

3|x-3|+2=14 |minus 2
3|x-3|=12 |divide 3
|x-3|=4

From absolute value definition you've got two ways:

x-3=4 or x-3=-4
x=7 or x=-1

And answer d) passes

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62°

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3 years ago

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1 year ago

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2 years ago

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Please help fast! give me the equation

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$\begin{bmatrix}x=\frac{\pi }{2}+2\pi n\\ x=\frac{3\pi }{2}+2\pi n\\ x=\frac{3\pi }{4}+\pi n\\ x=\frac{7\pi }{4}+\pi n\end{bmatrix}$

$\begin{bmatrix}\frac{3\pi }{2}+2\pi n=\frac{\pi }{2}+2\pi n\\ \frac{3\pi }{2}+2\pi n=\frac{3\pi }{4}+\pi n\\ \frac{3\pi }{2}+2\pi n=\frac{7\pi }{4}+\pi n\end{bmatrix}$

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2 years ago

Of the volunteers donating blood in a clinic, 30% have the Rhesus (Rh) factor present in their blood. (a) If five volunteers are

maksim [4K]

Answer:

0.83193

Step-by-step explanation:

Probability of having the Rhesus (Rh) factor present in their blood; p = 0.3

Sample size; n = 5

We want to find the probability the probability that at least one does not have the Rh factor.

This will be;1 - P(X < 1)

This is a binomial probability distribution. Thus;

P(X = k) = C(n, k) × p^(k) × (1 - p)^(n - k)

P(X < 1) = P(X = 0)

Thus;

P(X = 0) = C(5, 0) × 0.3^(0) × (1 - 0.3)^(5 - 0)

P(X = 0) = 1 × 1 × 0.16807

P(X = 0) = 0.16807

Thus;

1 - P(X < 1) = 1 - 0.16807 = 0.83193

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3 years ago