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garri49 [273]
3 years ago
8

I Need Help with math

Mathematics
1 answer:
Delicious77 [7]3 years ago
4 0
? Please elaborate on your problems so that others can help you prevail
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an employee has an annual salary of $48,700. they receive $1,530 in health insurance and $2,810 in paid time off per year. they
Anna [14]
Compensation for the year:$51,810(deducting vehicle expense)

An employee makes 48,700+1,530(health insurance)+2,810(paid time)=

53,040.  Also they receive 0.53 per mile x 9,000 miles=4,770

53,040 +4,770=57,810(employment compensation)

Personal work vehicle expense: 500 x 12 months=6,000

57,810 - 6,000= 51,810
6 0
3 years ago
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A man tied a rope to the top of a tree, which is 'x' m tall. The other end of the rope was tied to
Monica [59]

Answer:

(x+16)^2=20^2+x^2

Step-by-step explanation:

Given there was no bending of the tree, the Pitagorean Theory would help finding the value of the trees hight.

x^2+32x+16^2=400+x^2

32x=(20-16)×(20+16)

x=144/32

x=4m and 50cm

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3 years ago
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What is -10=10(x+9) and could you tell me how to do it cause I don't understand.​
timofeeve [1]
-10=10x+90
80=10x
x=80/10
x=8
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3 years ago
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The diagram shows a plan for fencing off a section of a public park to make a dog park. A wall that separates the park from a hi
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Where’s the diagram?
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3 years ago
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Historically 80% of the monitors made by your company will pass your stringent quality control checks. 40 monitors have just com
madam [21]

Answer:

a) 10.75% probability that exactly 30 of them will pass the quality control checks.

b) Expected value is 32

Variance is 6.4

Step-by-step explanation:

For each monitor, there are only two possible outcomes. Either they will pass the quality checks, or they will not pass these checks. The probability of a monitor passing these checks is independent of other monitors. So we use the binomial probability distribution to solve this question.

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 combinations 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.

The expected value of the binomial distribution is:

E(X) = np

The variance of the binomial distribution is:

V(X) = np(1-p)

80% of the monitors made by your company will pass your stringent quality control checks.

This means that p = 0.8

40 monitors have just come off the assembly line.

This means that n = 40

a) What is the probability that exactly 30 of them will pass the quality control checks?

This is P(X = 30). So

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

P(X = 30) = C_{40,30}.(0.8)^{30}.(0.2)^{10} = 0.1075

10.75% probability that exactly 30 of them will pass the quality control checks.

b) What are the expected value and variance of the number of these monitors that will pass the quality control checks?

E(X) = np = 40*0.8 = 32

Expected value is 32

V(X) = np(1-p) = 40*0.8*0.2 = 6.4

Variance is 6.4

4 0
3 years ago
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