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Zolol [24]
2 years ago
15

two companies rent moving trucks. pack and go charges $39.95 flat fee and $0.45 for every mile driven. U-drive charges $59.95 fl

at fee and $0.35 for every mile driven. At how many miles is the cost of the truck rentals the same
Mathematics
2 answers:
Nutka1998 [239]2 years ago
8 0

Answer:

$62.60

Step-by-step explanation:

C(190) = 22 + 0.1*190 = 22 + 19 = $41

C(406) = 22 + 0.1*406 = 22+40.6 = $62.60

zimovet [89]2 years ago
4 0

Answer:

Which equation represents the scenario?

39.95 + .45x = 59.95 + .35x

What does the variable x represents?  

X =miles driven

At how many miles is the cost of the truck rentals the same?

200

Step-by-step explanation:

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Dean is comparing prices on ground beef. Store A is selling 5 pounds of ground beef for $23.49. Store B is selling 8 pounds of g
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Answer: Store B

Step-by-step explanation:

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2 years ago
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Describe the transformation f(x) = √x + 1 − 5 in words. The parent function f(x) = √x has been shifted 1 unit and 5 units.
KATRIN_1 [288]

Answer:

The translation is 1 unit to the left and 5 units down

Step-by-step explanation:

we have

g(x)=\sqrt{x+1}-5

f(x)=\sqrt{x} -----> parent function

we know that

The transformation

f(x)-----> g(x) has the following rule

(x,y)-------> (x-1,y-5)

That means

The translation is 1 unit to the left and 5 units down

see the attached figure to better understand the problem

6 0
2 years ago
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In a study of the progeny of rabbits, Fibonacci (ca. 1170-ca. 1240) encountered the sequence now bearing his name. The sequence
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The question in part c is not clear, nevertheless, part a and part b would be solved.

Answer:

a. The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

b. The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

Step-by-step explanation:

a. Given

an + 2 = an + an + 1

where a1 = 1 and a2 = 1.

a3 = a1 + a2

= 2

a4 = a2 + a3

= 3

a5 = a3 + a4

= 5

a6 = a5 + a4

= 8

a7 = a6 + a5

= 13

a8 = a7 + a6

= 21

a9 = a8 + a7

= 34

a10 = a9 + a8

= 55

a11 = a10 + a9

= 89

a12 = a11 + a10

= 144

The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

(b)

Given

bn = an+1/an

b1 = a2/a1

= 1/1 = 1.000

b2 = a3/a2

= 2/1 = 1.000

b3 = a4/a3

= 3/2 = 1.500

b4 = a5/a4

= 5/3 = 1.667

b5 = a6/a5

= 8/5 = 1.600

b6 = a7/a6

= 13/8 = 1.625

b7 = a8/a7

= 21/13 = 1.615

b8 = a9/a8

= 34/21 = 1.619

b9 = a10/a9

= 55/34 = 1.618

b10 = a11/a10

= 89/55 = 1.618

The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

6 0
2 years ago
A quality control engineer tests the quality of produced computers. suppose that 5% of computers have defects, and defects occur
11111nata11111 [884]
(a) 0.059582148 probability of exactly 3 defective out of 20

 (b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.

  (a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So

 0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)

 = 0.05^3 * 0.95^17 * 20! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)

 = 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)

 = 0.05^3 * 0.95^17 * 20*19*3

 = 0.000125* 0.418120335 * 1140

 = 0.059582148

  (b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.

 The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)

 = 0.05*0.95^3*24/(1!3!)

 = 0.05*0.857375*24/6

 = 0.171475

 
 So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
3 0
2 years ago
In a health club, research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard dev
Paha777 [63]

Answer:

0.3114

Option d is right

Step-by-step explanation:

Let X be the time spent on a treadmill in the health club

Given that  research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard deviation of 5.4 minutes

Also given that X is normal

the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill.

= P(30

round off to two decimals tog et

the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill is 0.31

Hence option d is right

3 0
2 years ago
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