Ask question

Ask question

All categories

2 years ago

12

One serving of a cereal is 3/4 cup. Each box cereal has about 9 cups of cereal an costs $3.80. Is the price per serving greater

than $0.50? Explain

1 answer:

Zanzabum2 years ago

8 0

I think it is greater

You might be interested in

Solve this equation <br> 6. 14=t -44

Marizza181 [45]

Are you solving for t?

5 0

2 years ago

EleoNora [17]

Answer:

a.(-b) ( third choice)

explanation:

-a+b - there is no plus sign for -a.b.

not -a+b

(-a)(-6)=6a (-negative sign and -negative sign=+positive sign)

5 0

2 years ago

On a recent day, one U.S dollar was worth 0.81 euros. If you were to exchange 222 euros, how many U.S dollars would you receive?

Juli2301 [7.4K]

<em>Answer:</em>

<em>254.22</em>

<em>Step-by-step explanation:</em>

<em>222 euros equal 254.22</em>

7 0

2 years ago

A rectangle has a length four more than

torisob [31]

L ≥ 5w + 4
2(L+w) ≥ 32

L+w ≥ 16

6w + 4 ≥ 16
6w ≥ 12
w ≥ 2

L ≥ 5w + 4
L ≥ 14

If L = 14 and w = 2 then the perimeter is 32.
These are the smallest dimensions that fit the criteria.

Note: If we raise the width by 1 to w = 3
then the length is at least 19. At 19 by 3 we have P = 44
Like would be appreciated!!

8 0

8 months ago

A certain firm has plants A, B, and C producing respectively 35%, 15%, and 50% of the total output. The probabilities of a non-d

Sliva [168]

Answer:

There is a 44.12% probability that the defective product came from C.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

-In your problem, we have:

P(A) is the probability of the customer receiving a defective product. For this probability, we have:

$P(A) = P_{1} + P_{2} + P_{3}$

In which $P_{1}$ is the probability that the defective product was chosen from plant A(we have to consider the probability of plant A being chosen). So:

$P_{1} = 0.35*0.25 = 0.0875$

$P_{2}$ is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

$P_{2} = 0.15*0.05 = 0.0075$

$P_{3}$ is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

$P_{3} = 0.50*0.15 = 0.075$

So

$P(A) = 0.0875 + 0.0075 + 0.075 = 0.17$

P(B) is the probability the product chosen being C, that is 50% = 0.5.

P(A/B) is the probability of the product being defective, knowing that the plant chosen was C. So P(A/B) = 0.15.

So, the probability that the defective piece came from C is:

$P = \frac{0.5*0.15}{0.17} = 0.4412$

There is a 44.12% probability that the defective product came from C.

3 0

2 years ago