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mr_godi [17]
3 years ago
12

Ernie ate 2 1/3 slices of pepperoni pizza, 1 1/4 slices of sausage pizza, and 3 1/6 slices of cheese pizza for supper. How many

slices of pizza did he eat?
Mathematics
2 answers:
andrew11 [14]3 years ago
7 0

Answer:

The answer is 6 3/4

Step-by-step explanation:

Well first you are gonna have to extract the whole numbers in this question. the whole numbers are 3, 2, and 1. 3 + 2+ 1 is 6. Now we have to add the fractions. Since they all have different denominators we will have to find a multiple that they all have. all three of the denominators can be multiplied to get to 12. 1/6 times 2/2 = 2/12. 1/3 times 4/4 = 4/12. ¼ times 3/3 = 3/12. Now that we have the three denominators that are the same, we can add the fractions. 2/12 + 4/12 + 3/12 = 9/12. 9/12 simplified is 3/4. This means that the answer is 6 ¾.

Please mark me Brainliest if this helped! Thank you and have a nice day!

katrin [286]3 years ago
7 0

Answer:

6.75 slices

Step-by-step explanation:

To solve this question we will have to interprete

2 1/3 pepperoni pizza

1 1/4 sausage pizza

3 1/6 cheese pizza

So to determine the total we will have to add everything together

2 1/3+1 1/4+3 1/6

Let's convert them to improper fraction first

7/3+5/4+19/6

Let's find the LCM and the LCM is 12

28+15+38/12

81/12

Let's convert to proper fraction

6.75

Therefore he consumed 6.75 slices of pizza

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1814400

Step-by-step explanation:

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2 years ago
a student takes two subjects A and B. Know that the probability of passing subjects A and B is 0.8 and 0.7 respectively. If you
aniked [119]

Answer:

0.64 = 64% probability that the student passes both subjects.

0.86 = 86% probability that the student passes at least one of the two subjects

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Passing subject A

Event B: Passing subject B

The probability of passing subject A is 0.8.

This means that P(A) = 0.8

If you have passed subject A, the probability of passing subject B is 0.8.

This means that P(B|A) = 0.8

Find the probability that the student passes both subjects?

This is P(A \cap B). So

P(B|A) = \frac{P(A \cap B)}{P(A)}

P(A \cap B) = P(B|A)P(A) = 0.8*0.8 = 0.64

0.64 = 64% probability that the student passes both subjects.

Find the probability that the student passes at least one of the two subjects

This is:

p = P(A) + P(B) - P(A \cap B)

Considering P(B) = 0.7, we have that:

p = P(A) + P(B) - P(A \cap B) = 0.8 + 0.7 - 0.64 = 0.86

0.86 = 86% probability that the student passes at least one of the two subjects

3 0
3 years ago
Can anyone please help me out?
dem82 [27]
I think D is the answer
4 0
3 years ago
11 1/4-2 5/8= <br><br>then simplify
kkurt [141]
8.625 is the correct answer.
6 0
3 years ago
Read 2 more answers
Please help me if you know. Please give me an answer
Kryger [21]

9514 1404 393

Answer:

  6 3/10 pounds

Step-by-step explanation:

The weight change will be found by multiplying the rate of change by the time.

  ∆w = (-1.8 lb/h)(3.5 h) = -6.3 lb

The total change in weight after 3 1/2 hours is 6 3/10 = 6.3 pounds.

5 0
3 years ago
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