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Lubov Fominskaja [6]
3 years ago
13

The Quik-Clean Car Wash charges $15 for a car wash plus $2.50 per gallon for unleaded gasoline. The Mighty Clean Car Wash charge

s $10 for a carwash plus $2.75 per gallon for unleaded gasoline. Josh is comparing the cost of having his car wash plus filling up his tank with gas at each business. How many gallons of gas would he need to purchase for the cost to be the same at each car wash?
Mathematics
1 answer:
DerKrebs [107]3 years ago
3 0

Answer:

<em>Josh would need to purchase 20 gallons of gas for the cost to be the same at each car wash</em>

Step-by-step explanation:

We need to model both car wash's costs. The Quik-Clean Car Wash charges $15 for a car wash plus $2.50 per gallon for unleaded gasoline. If x is the number of gallons of unleaded gasoline. the total charges of this store are:

C1=15+2.5*x

The Mighty Clean Car Wash charges $10 for a carwash plus $2.75 per gallon for unleaded gasoline. Following the same procedure, their costs are:

C2=10+2.75*x

Josh wants to know how many gallons of gas will make both businesses charge the same. Equating both equations:

15+2.5*x=10+2.75*x

Simplifying:

-0.25x=-5

Solving:

x=-5/(-0.25)=20

Josh would need to purchase 20 gallons of gas for the cost to be the same at each car wash

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elena-14-01-66 [18.8K]

Answer:

5 ft

Step-by-step explanation:

Area of a rectangle

length x width

(3x-2) * (x) = 65

3x^2 - 2x - 65 = 0

Δ/4 = 1 + 195 = 196

x1 = (1 + 14)/3 = 5

x2 (1-14)/3 = -13/3 (not acceptable because a length can’t be negative)

3 0
3 years ago
Hello can someone try and help me
lyudmila [28]

Answer:

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4 0
3 years ago
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8 0
3 years ago
A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin
Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

P(\dfrac{X}{Y}) = 60\% = 0.6

P(\dfrac{X}{Y'}) = 0.01\% = 0.0001

P(Y) = 0.01

Thus, P(Y') \ will\ be = 1 - P(Y)

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

P(YX) = P(\dfrac{X}{Y}) \ P(Y)

P(YX) =0.6 \times 0.01

P(YX) =0.006

The probabilities of not involved in cheating & the evidence are present is:

P(Y'X) = P(Y')  \times P(\dfrac{X}{Y'})

P(Y'X) = 0.99  \times 0.0001 \\ \\  P(Y'X) = 0.000099

(b)

The required probability that the evidence is present is:

P(YX  or Y'X) = 0.006 + 0.000099

P(YX  or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}

P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}

P(\dfrac{Y}{X}) = 0.9838

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3 years ago
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Answer:

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