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Amiraneli [1.4K]
3 years ago
8

I need help with this question?????

Mathematics
1 answer:
inn [45]3 years ago
3 0
(5 cases)/(2 3/8 hour)
(5 cases)/(19/8 hours)
(5*8/19) cases/hour....invert the fraction and multiply 1/(19/8 hour)=(8/19)/hour
40/19 cases/hour
2 2/19 cases/hour
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Derek works at a lab with a huge circular particle accelerator. It has a diameter of 6 kilometers. What is the accelerator's rad
Rudiy27

Answer:

18.84

Step-by-step explanation:

Multiply the diameter by pi or 3.14

3 0
3 years ago
willie's school is selling tickets to the annual talent show. On the first day of ticket sales the school sold 10 adult tickets
JulsSmile [24]

Adult tickets - $15

Student tickets - $11

First day:

10 tickets * $15 = $150

4 tickets * $11 = $44

Total: $194

Second day:

7 tickets * $15 = $105

7 tickets * $11 = $77

Total: $182

7 0
2 years ago
Chelsea took out three loans for a total of ​$ 78 comma 000 78,000 to start an organic orchard. Her​ business-equipment loan was
lianna [129]

Answer:

Business-equipment loan, A=$15000

Small-business loan, B=$24000

home-equity loan, C= $39000

Step-by-step explanation:

Let the Business-equipment loan at an interest rate of​ 11%=A

Let the​ small-business loan was at an interest rate of 5​% =B

Let her​ home-equity loan was at an interest rate of 4.5​%.=C

Since her total Loan=$78000

A+B+C=78000....(I)

Simple Interest = (P X R X T)/100

Since the Time, T=1 year

Interest on A = 0.11A

Interest on B = 0.05B

Interest on C = 0.045C

The total simple interest due on the loans in one year was ​$4605.

0.11A+0.05B+0.045C=4605....(II)

The annual simple interest on the​ home-equity loan was ​$105 more than the interest on the​ business-equipment loan.

0.045C = 0.11A +105...(III)

We proceed to solve the simultaneous equations.

A+B+C=78000....(I)

0.11A+0.05B+0.045C=4605....(II)

0.045C = 0.11A +105...(III)

From (III), 0.045C = 0.11A +105

Substitute 0.045C = 0.11A +105 into (II).

0.11A+0.05B+0.11A +105=4605

0.22A+0.05B=4500

From (III),

C=\frac{22A}{9}+\frac{7000}{3}

Substitute into (I)

A+B+\frac{22A}{9}+\frac{7000}{3}=78000

\frac{31A}{9}+B=78000-\frac{7000}{3}

\frac{31A}{9}+B=\frac{227000}{3}

\frac{31A+9B}{9}=\frac{227000}{3}

3(31A+9B)=9 X 227000

31A+9B=681000

0.22A+0.05B=4500 (Multiply by 9)

31A+9B=681000 (Multiply by 0.05)

1.98A+0.45B=40500

1.55A+0.45B=34050

Subtracting

0.43A=6450

A=$15000

From (III)

C=\frac{22A}{9}+\frac{7000}{3}

C=\frac{22X15000}{9}+\frac{7000}{3} =39000

C=$39000

from (I)

A+B+C=78000

15000+B+39000=78000

B=$24000

8 0
3 years ago
Which of the following lines is perpendicular to the equation given below?
konstantin123 [22]

Answer:

C

Step-by-step explanation:

x+2y=8x...............

4 0
3 years ago
Read 2 more answers
Three workers arrive at work 15 minutes late, at a cost of $11/ hour.
PIT_PIT [208]
The cost for being late is $2.75 because 11 divided by four is 2.75
6 0
3 years ago
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