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Thepotemich [5.8K]
4 years ago
8

Harry has a small business cleaning kitchens and bathrooms. He usually cleans a bathroom in 1 hour and cleans a kitchen in 45 mi

nutes. He never works more than 15 hours in a week. Harry earns $60 per bathroom and $20 per kitchen job. He does not do more than 8 bathroom jobs per week (the smell gets to him). Find a combination of bathroom and kitchen jobs per week that will maximize his income and state the amount.
Mathematics
1 answer:
Bumek [7]4 years ago
4 0

Answer:

8 bathroom jobs and 9 kitchen jobs

Step-by-step explanation:

B=60

K=20

8*60=480

9*20=180

that would give harry 660 dollars in a week. HOWEVER- we have to make sure that its equal to or less than 15 hours of work.

8*1h= 8 hours in bathroom

9*45m=6.75hr in kitchen

8 hours+6.75 hours=14.75hr 14.75 hr<15hr so it works.

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Studentka2010 [4]
Alright, so for AB and CD to be parallel, CX and DX would have to be equal, as is with AX and BX. In addition, for CD and AB to be parallel, all sides in both triangles are either equal or not all sides in even one triangle are equal. Therefore, CD is not 3. In addition, two sides of a triangle combined must be greater than the third, so that leaves 5, 4, and 2 for CD. If it was 5, that would mean that all sides are equal, so that leaves 4 and 2. However, I don't see anything to prove either one right, sorry:/
8 0
3 years ago
Ahmed drive his car 4 hours continuously as average 80 Km/hr .if his speed was 60 Km in the 1st hour, 90 Km in the 2nd hour and
nadezda [96]

Answer:

I would say 120 but that could be also be wrong but ill go with 120

7 0
2 years ago
What is the value of the remainder if 10x4 – 6x3 + 5x2 – x + 1 is divided by x – 3?
natita [175]

Answer:

Remainder is 691.

Step-by-step explanation:

Given function is 10x^4-6x^3+5x^2-x+1.

Now we need to find remainder if we divide given function 10x^4-6x^3+5x^2-x+1 by (x-3)

(x-3) means plug x=3 into 10x^4-6x^3+5x^2-x+1 to find remainder.

10x^4-6x^3+5x^2-x+1

=10(3)^4-6(3)^3+5(3)^2-(3)+1

=10(81)-6(27)+5(9)-(3)+1

=810-162+45-3+1

=856-162-3

=856-165

=691

Hence remainder is 691.

8 0
3 years ago
Read 2 more answers
Factor x2 + 10x – 18. prime (x + 3)(x – 6) (x – 2)(x – 9) (x – 2)(x + 5)
Marta_Voda [28]

Answer:

prime

Step-by-step explanation:

x^2 + 10x – 18

What two numbers multiply to -18 and add to +10

There are no numbers that multiply to -18 and add to 10

-1 *18 = -18   -1 +18 = 17

-2 *9 = -18   -2 +9 = 7

This cannot be factored

8 0
3 years ago
For each set of probabilities, determine whether the events A and B are independent or dependent.
lozanna [386]

Answers:

  • (a) Independent
  • (b) Dependent
  • (c) Dependent
  • (d) Independent

========================================================

Explanation:

If events A and B are independent, then the two following equations must both be true

  • P(A | B) = P(A)
  • P(B | A) = P(B)

This is because the conditional probability P(A|B) means "P(A) when B has happened". If B were to happen, then P(A) must be the same as before. In other words, event B does not affect A, and vice versa.

For part (a), we have P(B) = 1/4 and P(B|A) = 1/4 showing that P(B|A) = P(B) is true, and therefore we can say the events are independent. We don't need the info that P(A) = 1/8.

------------------------

Unlike part (a), part (b) has the answer "dependent" because P(A) = 1/8 and P(A | B) = 1/3 differ in value. Event A starts off at probability 1/8, but then event B occurring means P(A) gets increased to 1/3. The prior knowledge about B changes the chances of A. The P(B) = 1/5 is unneeded.

------------------------

If A and B were independent, then,

P(A and B) = P(A)*P(B)

However,

P(A)*P(B) = (1/4)*(1/5) = 1/20

which is not the same as P(A and B) = 1/6. Therefore the two events are dependent.

------------------------

Refer back to part (a)

P(A) = 1/4 and P(A|B) = 1/4 are identical in value, so P(A|B) = P(A) which leads to the events being independent. Whether we know event B happened or not, it does not affect the outcome of event A. P(B) = 1/9 is unneeded.

7 0
2 years ago
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