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3 years ago

Given P(E)= 0.24 and P(FnE)= 0.17, find P(F| E). Round the answer to the nearest hundredth. A. 029 B. 0.40 C. 0.71 D. 1.41

Mathematics

1 answer:

romanna [79]3 years ago

7 0

Given P(E) = 0.24 and P(F ∩ E) = 0.17

It says to find conditional probability of F given E has occurred.

We know the formula of conditional probability is given by :-

P(F ║ E) = $\frac{P(E\;and\; F)}{P(E)}$

P(F ║ E) = $\frac{0.17}{0.24}$ = 0.708333

P(F ║ E) = 0.71

Hence, option C i.e. 0.71 is the final answer.

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The function of (x) varies directly with x^2, and f(x)=96 when x=4 what is the value of f(2)

makkiz [27]

$\bf \qquad \qquad \textit{direct proportional variation}
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\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby 
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\qquad variation
\end{array}\\\\
-------------------------------$

$\bf \textit{f(x) varies directly with }x^2\qquad f(x)=kx^2
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3 years ago

Just want to check to make sure this is right. I got 81 as the average so D seemed right.

sergiy2304 [10]

Im not 100% sure but im pretty sure you are right brother! God bless...Good luck!

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3 years ago

In a certain Algebra 2 class of 27 students, 18 of them play basketball and 16

Bogdan [553]

Answer:

<em>The probability is 0.89 or 89%</em>

Step-by-step explanation:

<u>Probability</u>

There are n=27 students in a certain Algebra 2 class. 18 out of them play basketball and 16 of them play baseball.

We also know 3 students play neither sport. This leaves 27-3=24 students playing sports.

The probability that a randomly chosen student plays basketball or baseball is:

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Simplifying:

$\displastyle p=\frac{8}{9}=0.89$

The probability is 0.89 or 89%

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Answer:

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Prime factorisation of 20 = 2×5×2

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Hence, the smallest number of each type will be 120.

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