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Elena-2011 [213]
3 years ago
8

Why does it help to rearrange

Mathematics
1 answer:
hichkok12 [17]3 years ago
8 0

Answer:

You don't really need to do it, but it helps you keep things more organized and easier to follow. Imagine if you're doing some multi-variable equation,

2a + 5b + 4d + 3c + b + a + 2d

that looks like a mess, it'll be easier to look at if you put all the similar variables next to each others like this:

a + 2a + b + 5b + 3c + 2d + 4d

(a + 2a) + (b + 5b) + 3c + (2d + 4d)

now you can add them up much easier.

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7groups

Step-by-step explanation:

15+6/3=7groups

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4−3(x−5)>6x−17 Given
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4 0
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Evaluate the expression for the given values.<br> 12x+5y/3z, where x + 1/2, y = 6 and z = 3
Semmy [17]

Answer:

4

Step-by-step explanation:

Substitute: \frac{\frac{12}{2} +5*6}{3*3}

Cross out the common factor: \frac{6+5*6}{3*3}

Calculate the product or quotient: \frac{6 +30}{3*3}

Calculate the product or quotient: \frac{6+30}{9}

Calculate the sum or difference: \frac{36}{9}

Cross out the common factor: 4

Answer: 4

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2 years ago
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Step-by-step explanation:

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3 years ago
Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

Z = \frac{X - \mu}{\sigma}

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

5 0
3 years ago
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