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Korolek [52]
2 years ago
7

Math how many models of 100 do u need to to model 3,200 explain

Mathematics
1 answer:
Keith_Richards [23]2 years ago
4 0
32 because 32 x 100 = 3,200
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Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu
Svetllana [295]

Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

a.

P(X>27)

First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of P(Z, you have to do the following conervertion:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.

c.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

d.

No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

4 0
2 years ago
Bob has 40 cents in his pocket. If Bob has no pennies, how many different combinations of quarters, dimes, and/or nickles does h
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Answer:

He could have 4 dimes, or 8 nickles, or 1 quater and 3 nickels, or 1 quater with one dime and 1 nickel, or 3 dimes and 2 nickles, or 1 quater with obe nickel and one dime.

Step-by-step explanation:

hope this helps.

3 0
2 years ago
Suppose we want to choose 2 objects, without replacement, from the 5 objects pencil, eraser, desk, chair, and lamp. (a)How many
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Answer:

a) 20 ways

b) 10 ways

Step-by-step explanation:

When the order of selection/choice matters, we use Permutations to find the number of ways and if the order of selection/choice does not matter, we use Combinations to find the number of ways.

Part a)

We have to chose 2 objects from a group of 5 objects and order of choice matters. This is a problem of permutations, so we have to find 5P2

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Therefore, we can choose 2 objects from a group of 5 given objects if the order of choice matters.

Part b)

Order of choice does not matter in this case, so we will use combinations to find the number of ways of choosing 2 objects from a group of 5 objects which is represented by 5C2.

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Using the value of n=5 and r=2, we get:

5C2=\frac{5!}{2!(5-2)!} =10

Therefore, we can choose 2 objects from a group of 5 given objects if the order of choice does not matters.

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2 years ago
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The correct answer is choice D.  If you put choice D into words, it is saying 30% (0.3) of the total is 12 minutes.

To solve this, use inverse operations.

<span><u>0.3m</u> = <u>12</u>
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