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Delvig [45]
3 years ago
7

Fathi has $1.10 , in his printing account. Each sheet of paper he uses reduces his printing account balance by $0.25. Fathi want

s to print out a PDF document that is 47 pages long. To save paper, he decides to print on both sides of each sheet and to print two pages on each side of the sheet.
After Fathi prints, what will be the balance in his printing account?
Mathematics
1 answer:
lianna [129]3 years ago
3 0

Answer:

-$1.9

Step-by-step explanation:

There are 47 pages.

Printing on both sides would divide the number of pages into half.

47/2 = 23.5

2 pages on each side would mean 4 pages on one sheet. Therefore, the number of pages will be further divided by 2.

23.5/2 = 11.75

There cannot be 11.75 pages so we will round it up to 12 pages.

Each page costs $0.25 so 12 pages will cost:

12 x 0.25 = $3

Faithi has $1.1 so new account balance will be:

1.1 - 3 = $-1.9

Therefore, Fathi's balance in his printing account would be negative $1.9.

!!

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To find the volume you use the formula

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A fair coin is tossed three times and the events A, B, and C are defined as follows: A: \{ At least one head is observed \} B: \
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Answer:

a) P(A)=0.875

b) \text{P(A or B)}=0.875

c) \text{P((not A)  or B  or (not C))}=0.625

Step-by-step explanation:

Given : A fair coin is tossed three times and the events A, B, and C are defined as follows: A: At least one head is observed, B: At least two heads are observed, C: The number of heads observed is odd.

To find : The following probabilities by summing the probabilities of the appropriate sample points ?

Solution :

The sample space is

S={HHH,HHT,HTT,HTH,TTT,TTH,THH,THT}

n(S)=8

A: At least one head is observed

i.e. A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

n(A)=7

B: At least two heads are observed

i.e. B={HHH,HTT,TTH,THT}

n(B)=4

C: The number of heads observed is odd.

i.e. C={HHH,HTT,THT,TTH}

n(c)=4

a) Probability of A, P(A)

P(A)=\frac{n(A)}{n(S)}

P(A)=\frac{7}{8}

P(A)=0.875

b) P(A or B)

Using formula,

\text{P(A or B)}=P(A)+P(B)-\text{P(A and B)}

\text{P(A or B)}=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{\text{n(A and B)}}{n(S)}

\text{P(A or B)}=\frac{7}{8}+\frac{4}{8}-\frac{4}{8}

\text{P(A or B)}=\frac{7}{8}

\text{P(A or B)}=0.875

(c) P((not A)  or B  or (not C))

A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

not A = {TTT} = 1

B={HHH,HTT,TTH,THT}

C={HHH,HTT,THT,TTH}

not C = {HHT,HTH,THH,TTT} = 4

So, not A or B or not C = {HHH,HHT,HTH,THH,TTT}=5

\text{P((not A)  or B  or (not C))}=\frac{5}{8}

\text{P((not A)  or B  or (not C))}=0.625

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L (water level) = -0.5 d (days) +34

Place 26 into the (L) variable and solve

26= -0.5 d + 34

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d=16
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