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2 years ago

15

Two mechanics worked on a car. The first mechanic worked for 15 hours, and the second mechanic worked for 5 hours. Together they

charged a total of $1075. What was the rate charged per hour by each mechanic if the sum of the two rates was $105 per hour?

1 answer:

leva [86]2 years ago

8 0

Now this may or may not be work I’m 99.9 percent sure it’s right but if not sorry

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Help me on the Elizabeth problem please

EastWind [94]

4 big mice and 3 small mice because 4 X 25=100; 3 X 10=30

100+30=130

is that right?

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2 years ago

There are 203 apples in a basket. How many children can share these apples equally? How many apples would each child get?

tangare [24]

I think this is factorials
so an example would b....... 1*2*3*4*5.... ect

hope this helps

6 0

3 years ago

Read 2 more answers

Christais attending the county fair that charges a $12 entry fee. The entry fee includes 10 free rides, but any additional rides

ratelena [41]

Answer:

he saves 3 dollars. its not worth it.

Step-by-step explanation:

you divide 12 by ten getting you 1.2 which you subtract from 1.50 that then multiply that by 10

8 0

2 years ago

Joe can cut and split a cord of firewood in 3 fewer hours than Dwight can. When they work together, it takes them 2 hours. How

aalyn [17]

Answer:

Dwight will take 6 hours to finish the job alone and Joe will take 3 hours to finish the job alone.

Step-by-step explanation:

Let us assume the time taken by Dwight to split a cord of firewood = K hrs

So, the per hour rate of Dwight = $(\frac{1}{K})$

As, Joe uses 3 LESS hours then Dwight.

So, the time taken by Joe to split a cord of firewood = (K- 3) hrs

So, the per hour rate of Joe = $(\frac{1}{K-3})$

Now, when both of them wok together, it takes them 2 hours.

So, the per hour rate of BOTH of them = $(\frac{1}{2})$

⇒ Per hour rate of ( Dwight + Joe) = $(\frac{1}{2} )$

$\implies (\frac{1}{K}) + (\frac{1}{K-3}) = (\frac{1}{2})$

Now, solving for the value of K , we get:

$(\frac{1}{K}) + (\frac{1}{K-3}) = (\frac{1}{2})\\\implies \frac{(K-3) + K}{K (K-3)} = (\frac{1}{2})\\\implies 2(2K -3) = K^2 - 3K\\\implies k^2 - 3K -4K +6 = 0\\\implies K^2 - 7K + 6= 0\\\implies K^2 - 6K - K + 6= 0\\\implies K(K-6) -1(K - 6)= 0\\\implies (K-6)(K-1) = 0$

Implies either K = 6 Or K = 1

But if K = 1, (K-3) = 1- 3 = -2 hours would be A CONTRADICTION.

⇒ K = 6 hours

Hence, Dwight will take 6 hours to finish the job alone and Joe will take (k-3) = (6-3) = 3 hours to finish the job alone.

8 0

2 years ago

You own a portfolio that is invested 43 percent in stock a, 16 percent in stock b, and the remainder in stock

docker41 [41]

Multiply the investment fraction by the corresponding return, and add the products. The given fractions add to 59%, so the remainder is 41%. (The total of the fractions is 100%—all of the portfolio.)

... 0.43*9.1% + 0.16*16.7% + 0.41*11.4%

... = 3.913% + 2.672% + 4.674% = 11.259%

6 0

2 years ago