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just olya [345]
3 years ago
14

In a factory, 3.25% of the jars of peanut butter produced contains less peanut butter than is advertised on the label. The facto

ry produces 1200 jars of peanut butter every hour. In three 8- hour work days, how many jars of peanut butter do you expect to contain less peanut butter than usual?
Mathematics
1 answer:
marshall27 [118]3 years ago
7 0

Answer:

936 jars of peanut butter contain less peanut butter than usual.

Step-by-step explanation:

1. First, we need to find how many jars of peanut butter that contain less peanut butter than advertised are produced every hour.

The factory produces 1200 jars of peanut butter every hour

3.25% of those jars have less peanut butter than advertised

So: multiply (1200*.0325) (the decimal form of the percent) to find the amount of peanut butter jars that contain less peanut butter than advertised.

(1200*.0325) = 39

So the factory produces 39 jars of peanut butter that contain less peanut butter than advertised.

2. Second, we need to find how many jars of peanut butter that contain less peanut butter than advertised in three 8 hour work days.

Lets add these work days together to find the total amount of hours: (8 + 8 + 8) = 24

So: three 8 hour work days totals to 24 hours

Since the factory produces 39 jars of peanut butter that contain less peanut butter than advertised per hour. Multiply (24 * 39) to find the total amount of jars produced in three 8 hour work days.

(24 *39) = 936

A total of 936 peanut butter jars are produced in three 8 hour work days that contain less peanut butter than advertised.

3. If you want to double check your work do the following.

Since the factory produces 1200 jars an hour and 24 hours total three 8 hour work days. Multiply (1200*24) = 28800

So there is a total of 28800 jars produced in three 8 hour work days.

Then you can multiply 28800 by 3.25% of jars that contain less peanut butter to find the total amount of peanut butter jars that contain less peanut butter in three 8 hour work days.

(28800 * .0325) = 936

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Answer:

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Step-by-step explanation:

Let's start writing the sample space for this experiment :

S= { (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) }

Let's also define the event E ⇒

E : '' The sum of the two dice is 5 ''

We can describe the event by listing all the favorables cases from S ⇒

E = { (4,1) , (3,2) , (2,3) , (1,4) }

In order to calculate P(E) we are going to divide all the cases favorables to E over the total cases from S. We can do this because all 36 of these possible outcomes from S are equally likely. ⇒

P(E)=\frac{4}{36}=\frac{1}{9} ⇒

P(E)=\frac{1}{9}

Finally we are going to define the event F ⇒

F : '' The number of the first die is exactly 1 more than the number on the second die ''

⇒

F = { (2,1) , (3,2) , (4,3) , (5,4) , (6,5) }

Now given two events A and B ⇒

P ( A ∩ B ) = P(A,B)

We define the conditional probability as

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We need to find P(F|E) therefore we can apply the conditional probability equation :

P(F|E)=\frac{P(F,E)}{P(E)}   (I)

We calculate P(E)=\frac{1}{9} at the beginning of the question. We only need P(F,E).

Looking at the sets E and F we find that (3,2) is the unique result which is in both sets. Therefore is 1 result over the 36 possible results. ⇒

P(F,E)=\frac{1}{36}

Replacing both probabilities calculated in (I) :

P(F|E)=\frac{P(F,E)}{P(E)}=\frac{\frac{1}{36}}{\frac{1}{9}}=\frac{1}{4}=0.25

We find out that P(F|E)=\frac{1}{4}=0.25

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