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3 years ago

7

during the first round of game show, jackson scored 300 points. After the second round,after the second round, his total score w

as -300. how many points did jackson score in the second round? (it is not 300)

1 answer:

mrs_skeptik [129]3 years ago

4 0

Answer:

-600

Step-by-step explanation:

Jackson's initial score was 300, then his later score became -300, so you know that he had to lose points.

-300=300+x

** x represents the point change

Then just solve the equation by getting x by itself. You do this by subtracting 300 from both sides:

-300=300+x

-300 -300 = -300-300+x

-600 = x

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A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historic

stellarik [79]

Answer:

a) There is a 59.87% probability that none of the LED light bulbs are defective.

b) There is a 31.51% probability that exactly one of the light bulbs is defective.

c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) There is a 100% probability that three or more of the LED light bulbs are not defective.

Step-by-step explanation:

For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 10, p = 0.05$

a) None of the LED light bulbs are defective?

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}*(0.05)^{0}*(0.95)^{10} = 0.5987$

There is a 59.87% probability that none of the LED light bulbs are defective.

b) Exactly one of the LED light bulbs is defective?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{10,1}*(0.05)^{1}*(0.95)^{9} = 0.3151$

There is a 31.51% probability that exactly one of the light bulbs is defective.

c) Two or fewer of the LED light bulbs are defective?

This is

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 2) = C_{10,2}*(0.05)^{2}*(0.95)^{8} = 0.0746$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884$

There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) Three or more of the LED light bulbs are not defective?

Now we use p = 0.95.

Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.

So

$P(X \leq 2) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 0) = C_{10,0}*(0.95)^{0}*(0.05)^{10}\cong 0$

$P(X = 1) = C_{10,1}*(0.95)^{1}*(0.05)^{9} \cong 0$

$P(X = 2) = C_{10,1}*(0.95)^{2}*(0.05)^{8} \cong 0$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0$

$P(X \geq 3) = 1 - P(X \leq 2) = 1$

There is a 100% probability that three or more of the LED light bulbs are not defective.

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Suppose a car rental company charges $88 for the first day and $38 for each additional or partial day. Let S(x) represent the co

riadik2000 [5.3K]

Answer:

C. $240 for 4.5 days.

Step-by-step explanation:

Let's find the answer using the following data:

Notice that the first day is $88 which can be written as:

first day= $38 + $50, in this way the first day cost the same as the other additional days but it include an additional $50 charge.

In this way we can have the formula:

S(x)=38x+50

Notice in the above formula that the '50' represents the extra money for the first day, while '38' the normal cost per day.

Now, S(4.5) needs to be evaluated as S(5) given the condition that a partial day also costs $38, so:

S(4.5)=S(5)=(38*5)+50=240

In conclusion, for 4.5 days the cost is $240, so de answer is C.

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Alex paid a total of $530 for t tickets to a concert. How much did each ticket cost?

neonofarm [45]

Answer:

$530/T

Step-by-step explanation:

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