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Orlov [11]
3 years ago
9

William bought pencil boxes for his art class. Each box costs $3.50. He gave $30 to the cashier and got $5.50 as change. How man

y pencil boxes did he buy?
A. 3

B. 4

C. 7

D. 10


Help It's for benchmark testing so I need it now!! D:
Mathematics
1 answer:
mr_godi [17]3 years ago
8 0

Answer:

D

Step-by-step explanation:

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1.The school library had a used book sale. Paperback books were 25 cents each, and hardcover books were $1.25 each. There were 5
KatRina [158]
.25x54=13.5   163x1.25=203.75   13.5+203.75=217.25
4 0
3 years ago
Which equation has a solution of X=<br> 13?<br> O<br> X+ 3 = 10<br> O x-10=3<br> * = 26<br> Ox+12= 1
Marianna [84]

Answer:

x-10=3

x+10-10=3+10

x=13

3 0
3 years ago
Read 2 more answers
One number is 7 times another number. The product of the two numbers is 252. Find the two numbers.
Nitella [24]

Step-by-step explanation:

x =7y ----(1)

xy=252 ----(2)

(1) into (2):

(7y)(y) = 252

7 {y}^{2}  = 252 \\ 7y =  \sqrt{252}

7y=15.874....

y=2.27

y value into (1)

x= 7(2.268)

x=15.87

the two numbers are 15.87 and 2.27 (Ans)

6 0
3 years ago
I need help ASAP! Can anyone please check my work?
STALIN [3.7K]

A = event the person got the class they wanted

B = event the person is on the honor roll

P(A) = (number who got the class they wanted)/(number total)

P(A) = 379/500

P(A) = 0.758

There's a 75.8% chance someone will get the class they want

Let's see if being on the honor roll changes the probability we just found

So we want to compute P(A | B). If it is equal to P(A), then being on the honor roll does not change P(A).

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

A and B = someone got the class they want and they're on the honor roll

P(A and B) = 64/500

P(A and B) = 0.128

P(B) = 144/500

P(B) = 0.288

P(A | B) = P(A and B)/P(B)

P(A | B) = 0.128/0.288

P(A | B) = 0.44 approximately

This is what you have shown in your steps. This means if we know the person is on the honor roll, then they have a 44% chance of getting the class they want.

Those on the honor roll are at a disadvantage to getting their requested class. Perhaps the thinking is that the honor roll students can handle harder or less popular teachers.

Regardless of motivations, being on the honor roll changes the probability of getting the class you want. So Alex is correct in thinking the honor roll students have a disadvantage. Everything would be fair if P(A | B) = P(A) showing that events A and B are independent. That is not the case here so the events are linked somehow.

8 0
4 years ago
In an apartment complex with 28 units, 19 of the renters keep a pet. What percentage does not keep a pet?
Aleks04 [339]

If 19 out of 28 renters keep a pet, there are 28-19 = 9 renters who don't keep a pet.

Whenever you have a subset of some set, and you want to know which percentage of the set the subset represents, you simply have to compute

\dfrac{\text{number of elements in the subset}}{\text{number of elements in the set}}\times 100

So, in your case, you're wondering what percentage of 28 does 9 represent. So, the formula becomes

\dfrac{9}{28}\times 100 = 0.32\overline{142857}\times 100 = 32.\overline{142857} \approx 32.14\%

6 0
3 years ago
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