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3 years ago

Tony buys 2 boxes of chocolates. One is 4 in. 7 in. 8 in. And 8 in. 5 in. 6 in. Which box has the larger volume?

Mathematics

1 answer:

Rufina [12.5K]3 years ago

4 0

Answer:

Second box

Step-by-step explanation:

Volume (of the box) = length * width * height

Volume of first box is: 4 * 7 * 8 = 224

Volume of second box is: 8 * 5 * 6 = 240

Since 224 < 240, the volume of the second box has the larger value.

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The formula for calculating the future value of an investment:

FV = P (1 + r)^nm

FV = Future value
P = Present value
R = interest rate
m = number of compounding
N = number of years

Future value of Larry's investment: $350 x [1 + (0.04/4)]^(4 x 2) = $379

Future value of Peggy's investment: $350 x [1 + (0.06/12)]^(12 x 2) = $394.51

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2 years ago

Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper c

aalyn [17]

Answer:

a) This means that there is a 33% probability that one car chosen at random will have less than 49.5 tons of coal.

b) There is a 0.04% .probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The actual weights of coal loaded into each car are normally distributed, with mean = 50 tons and standard deviation = 0.9 ton. This means that $\mu = 50, \sigma = 0.9$

(a) What is the probability that one car chosen at random will have less than 49.5 tons of coal? (Round your answer to four decimal places.)

This is the pvalue of Z when $X = 49.5$

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

$Z = \frac{49.5 - 50}{0.9}$

$Z = -0.44$

$Z = -0.44$ has a pvalue of 0.33.

This means that there is a 33% probability that one car chosen at random will have less than 49.5 tons of coal.

(b) What is the probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal? (Round your answer to four decimal places.)

Now we have to use the standard deviation of the sample, since we are working with the sample mean. That is

$s = \frac{\sigma}{\sqrt{n}} = \frac{0.9}{\sqrt{35}} = 0.15$

Now, we find pvalue of Z when $X = 49.5$

$Z = \frac{X - \mu}{s}$

$Z = \frac{49.5 - 50}{0.15}$

$Z = -3.33$

$Z = -3.33$ has a pvalue of 0.0004.

This means that there is a 0.04% .probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal

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3 years ago