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klemol [59]
3 years ago
13

After a hike, a group of students equally share 5 boxes of granola bars. Each box has 8 granola bars. Which algebraic expression

represents this situation?
A. (5 × 8) ÷ s
B. 5 × (8 ÷ s)
C. s × (5 + 8)
D. (s + 5) × 8
Mathematics
1 answer:
SIZIF [17.4K]3 years ago
8 0

Answer: B.

First, we have to have a question for this problem to create a solution. So, the example sentence is how many granola bars does each student get.

Second, an algebraic expression has a variable in it. So since we are trying to find the number of students, the variable is s.

From looking at C and D, we can already eliminate them. This question has mostly multiplication and division words in it, so it can't be addition and multiplication.

Now, we only have A and B left. A tip on  how to figure this out is reading the word problem out thoroughly. I will use B as an example. 5 boxes of granola bars were multiplied by the number of granola bars each student got. From reading this, it makes more sense. But it does change the example sentence, though.

I hope this helped and this is the right answer!

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Which 37/99 as a decimal <br> A) 0.037<br> B) 0.37<br> C) 0.37<br> D) 3.73
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Step-by-step explanation:

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– StartFraction 5 Over 3 EndFraction v plus 4 equals 8 minus StartFraction 1 Over 3 EndFraction v.(6x – 3) = –
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6 0
4 years ago
A computer maker receives parts from three suppliers, S1, S2, and S3. Fifty percent come from S1, twenty percent from S2, and th
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Answer:

a) 4.9 % of all parts is defective or 0.049 of the total parts.

b)  0.5102 is the probability that the defective part was supplied by S1

Step-by-step explanation:

N is the total number of parts from supplier S1, S2 and S3.

N1 = 0.5*N is the total number of part supplied by S1

N2 = 0.2*N is the total number of part supplied by S2  

N3 = 0.3*N is the total number of part supplied by S3

a) if Nd1 is the number of defective parts from supplier S1, Nd2 is the number of defective parts from supplier S2 and Nd3 is the number of defective parts from supplier S3, the the total defective parts Nd is:

Nd = Nd1 + Nd2 + Nd3, where

Nd1 = 0.05*N1 = 0.05*0.5*N = 0.025*N,

Nd2 = 0.03*N2 = 0.03*0.2*N = 0.006*N,

Nd3 = 0.06*N3 = 0.06*0.3*N = 0.018*N,

Then Nd = Nd1 + Nd2 + Nd3 =  0.049*N, so Nd/N = 0.049

b) P(S1 \vert d) = \frac{P(S1,d)}{P(d)} = \frac{P(d \vert S1)}{P(d)} = \frac{0.05*0.5}{0.049} \approx 0.5102

for the last expression I used the Bayes tehorem.

P(S1 \vert d) is the probability that occur S1 given that d (defective) is true. This a conditional probability.

see at https://en.wikipedia.org/wiki/Bayes%27_theorem

6 0
3 years ago
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