Help ASAP I'm bad at solving this I will give out 20 points!

Mathematics

1 answer:

nexus9112 [7]4 years ago

3 0

Answer:

(0,3/2), (0,2),(6,0),(9/4,0)

I hope it helpss..

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A factory makes car batteries. The probability that a battery is defective is 1/14. If 400 batteries are tested, about how many

Rufina [12.5K]

Answer:

28.6, that is, about 29 are expected to be defective

Step-by-step explanation:

For each battery, there are only two possible outcomes. Either it is defective, or it is not. The probability of a battery being defective is independent of other betteries. So the binomial probability distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

The probability that a battery is defective is 1/14.

This means that $p = \frac{1}{14}$

400 batteries.

This means that $n = 400$

How many are expected to be defective?

$E(X) = np = 400*\frac{1}{14} = 28.6$

28.6, that is, about 29 are expected to be defective

3 0

3 years ago

Solve -3/4(8k-24)>-12

lisabon 2012 [21]

$-\dfrac{3}{4}(8k-24) > -12\ \ \ \ |\teext{change the signs}\\\\\dfrac{3}{4}(8k-24) < 12\ \ \ \ |\cdot\dfrac{4}{3}\\\\8k-24 < 12\cdot\dfrac{4}{3}\\\\8k-24 < 4\cdot4\\\\8k-24 < 16\ \ \ \ |+24\\\\8k < 40\ \ \ \ |:8\\\\k < 5\to k\in(-\infty;\ 5)$