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posledela
3 years ago
11

Pepper jackie has 12 picks 1 out of every 4 picks is orange the rest are green how many picks are orange

Mathematics
2 answers:
inessss [21]3 years ago
6 0
3 picks will be orange. 12/4 = 3
Len [333]3 years ago
3 0

Answer:

3

Step-by-step explanation:

Pepper Jackie has 12 picks, 1 out of every 4 pick is orange and the rest are green.

So there could be three groups of 4 each and in each group there is 1 orange.

There are 3 groups of 4 each that makes 12 (3 \times 4=12)

Now, in each group there is 1 orange, so in three groups there will be

1+1+1=3 orange.

1 for each group.

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Ilya [14]

Answer:

40%

Step-by-step explanation:

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Dishwashers: 1/20 = 5%

Dryers: 11/20 = 55%

So there is a 40% chance she will sell another refrigerator! hope this helps!!

3 0
2 years ago
You roll three die. what is the probability that 5 will not appear on the die?
castortr0y [4]
No of out comes =216
favorable outcome=3 
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7 0
3 years ago
Solve for a<br>-7=-ax-16​
Rudik [331]

Answer:

x=-9/a and a=-9/x

Step-by-step explanation:

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2 years ago
Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of
Alexxx [7]

Answer:

a) 0.018

b) 0            

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)

= 1 - P(z \leq 2.1)

Calculation the value from standard normal z table, we have,  

P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%

0.018 is the probability that if an individual man is randomly​ selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected​ men, these men have a mean hip breadth greater than 16.5 in)

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

P(x > 16.5)  

P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)  

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Calculation the value from standard normal z table, we have,  

P(x > 16.5) = 1 - 1 = 0

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in

4 0
3 years ago
Suppose the population of deer in a state is 15,430 and is growing 2% each year. Predict the population after 8 years.
Dennis_Churaev [7]
A = 15,430(1+.02)^8
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