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Gemiola [76]
2 years ago
8

The Yert family hires a landscaping company for 30 weeks. There are two companies to choose from. Green Lawns charges 1 for the

first time they service your lawn, and the price will multiply by a factor of 0.90 every week thereafter. Green Thumbs charges 400 for the first time they service your lawn, and then charges an additional 45 every week thereafter. At 30 weeks, if the Yert family chooses the company that is cheaper, how much money will they save?
Mathematics
1 answer:
zhuklara [117]2 years ago
6 0
Do you need help or u just wanted the answer
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Which of the following equations have exactly one solution?
Tems11 [23]

Answer:

A, B, & C

Step-by-step explanation:

A: -5x +12 = -12x - 12

    +12x        +12x

    7x + 12 = -12

         -12     -12

            7x = -24                 [Divide both sides by 7 to get x]

              x = -24/7

B: −5x + 12 = 5x + 12

   -5x            -5x

    -10x + 12 = 12

            -12     -12

            -10x = 0                   [Divide both sides by -10 to get x]

                 x = 0

C: −5x + 12 = 5x − 5

    -5x           -5x

    -10x + 12 = -5

            -12      -12

           -10x = -17                  [Divide both sides by -10 to get x]

               x = 17 / 10

D: −5x + 12 = −5x − 12

    +5x           +5x

              12 ≠ -12         [The statement is false, so it isn't the correct answer]

8 0
3 years ago
Use RACE strategy to solve the problem “Class Picnic”:
Ber [7]

Answer:

I thank that Jordan will buy 5 packs of hamburgers and he would buy 3 bags of hamburger buns. And each student can have 1 each.

Step-by-step explanation:

sorry if its wong

8 0
2 years ago
If you add one third of a number to the number itself you get 16 what is the number
sineoko [7]
The number is 12

12 + 4 = 16

4 is a third of 12 and 12 + 4 = 16.
6 0
3 years ago
The ratio of lemon to water in 80 fluid ounces of lemonade id 3:1 the lemonade was diluted futher if the ratio was 3:2 how much
andriy [413]

Answer:

I think 53.334 ounces were added to dilute the lemonade

4 0
2 years ago
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.

8 0
3 years ago
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