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2 years ago

What is the total weight in ounces of the three heaviest boxes?

Mathematics

2 answers:

kozerog [31]2 years ago

6 0

Answer:

D. 252 ounces

Step-by-step explanation:

The number line is cut in fourths. The 3 heaviest boxes are, 1 between 4 and 5 lbs, and 2 boxes between 5 and 6.

1st box: 4.75 lbs

2nd box: 5.5 lbs

3rd box: 5.5 lbs

Total lbs: 15.75 lbs

15.75 pounds= 252 ounces

Tcecarenko [31]2 years ago

4 0

D! Hope this helps :)

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What are the terms in the expression 2x – 4y + 8?

WITCHER [35]

2x - 4y + 8

The terns:

2x; -4y and 8

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3 years ago

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Consider the equation y = -2x + 5. Create a table of five ordered pairs that satisfy the equation. What is the y-intercept of th

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Answer: option d.

Step-by-step explanation:

To create a table of five ordered pairs that satisfy the equation, you need to give values to the variable "x" and substitute these into the function to find the corresponding value of "y":

For $x=-2$

$y = -2(-2) + 5=9$

For $x=-1$

$y = -2(-1) + 5=7$

For $x=0$

$y = -2(0) + 5=5$

For $x=1$

$y = -2(1) + 5=3$

For $x=2$

$y = -2(2) + 5=1$

Then, you can make the table:

x -2 -1 0 1 2

y 9 7 5 3 1

To find the x-intercept, substitute $y=0$ into the function and solve for "x". Then:

$0 = -2x + 5\\\\x=\frac{-5}{-2}\\\\x=\frac{5}2}$

This is: ( $\frac{5}{2}$ , 0)

To find the y-intercept, substitute $x=0$ into the function and solve for "y". Then:

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3 years ago

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Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Natali [406]

Answer:

The percentage of snails that take more than 60 hours to finish is 4.75%

The relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

There is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

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

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

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.

In this problem, we have that:

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean 50 hours and standard deviation 6 hours.

This means that $\mu = 50$ and $\sigma = 6$ .

The percentage of snails that take more than 60 hours to finish is %

The pvalue of the zscore of $X = 60$ is the percentage of snails that take LESS than 60 hours to finish. So the percentage of snails that take more than 60 hours to finish is 100% substracted by this pvalue.

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

A Zscore of 1.67 has a pvalue of .9525. This means that there is a 95.25% of the snails take less than 60 hours to finish.

The percentage of snails that take more than 60 hours to finish is 100%-95.25% = 4.75%.

The relative frequency of snails that take less than 60 hours to finish is

The relative frequence off snails that take less than 60 hours to finish is the pvalue of the zscore of X = 60.

In the item above, we find that this value is .9525.

So, the relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is:

This is the pvalue of the zscore of X = 67 subtracted by the pvalue of the zscore of X = 60. So

$X = 67$

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

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

A zscore of 2.83 has a pvalue of .9977.

For X = 60, we have found a Zscore o 1.67 with a pvalue of .9977

So, the percentage of snails that take between 60 and 67 hours to finish is:

$p = .9977 - 0.9525 = .0452$

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 100% subtracted by the pvalue of the Zscore of X = 76.

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

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The pvalue of Z = 4.33 is 1.

So, there is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most hours.

The most hours that a snail must finish is the value of X of the Zscore when p = 0.10.

Z = -1.29 has a pvalue of 0.0985, this is the largest pvalue below 0.1. So what is the value of X when Z = -1.29?

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

$-1.29 = \frac{X - 50}{6}$

$X - 50 = -7.74$

$X = 42.26$

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical 80% of snails take between and hours to finish.

This is from a pvalue of .1 to a pvalue of .9.

When the pvalue is .1, X = 42.26.

A zscore of 1.28 is the largest with a pvalue below .9. So

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

$1.28 = \frac{X - 50}{6}$

$X - 50 = 7.68$

$X = 57.68$

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

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alexandr1967 [171]

burned: 756 calories

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steps

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