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3 years ago

5

helen didvided her colored pencils evenly among herself and three freinds. write an expression to represent the number of pencil

s each person received

1 answer:

svetoff [14.1K]3 years ago

8 0

So, As Helen divides these pencils, she will divided among, herself, and also her other friends.

This would then signify that the expression would look like the following:

$\boxed{p\div(4)}$

We really do not know how many pencils Helen had, so this is why <em>p </em>is in the equation.

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PLEASE HELP 100 POINTS!!

tresset_1 [31]

y = x^2 + 5x -3

y-x =2

Replace y in the second equation with the first one:

x^2 + 5x -3 - x = 2

Simplify by combining like terms:

x^2 + 4x -3 = 2

Subtract 2 from both sides:

x^2 + 4x -5 = 0

Fctor the polynomial:

(x-1) (x+5) = 0

Solve for both X's:

x = 1 and x = -5

Now replace x in the second equation with both value and solve for y:

y -1 =2, y = 3

y - -5 =2, y +5 = 2, y = -3

Now combine the sets of answers:

(1,3) and (-5,-3)

3 0

4 years ago

Read 2 more answers

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.

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

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

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.

5 0

3 years ago

Christina made 84% of all shots attempted and mallory made 7/8 of all shots who made more

Ghella [55]

Hey!

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~Turn both into decimals.

84% = 0.84

7/8 = 0.875

--------------------------------------------------

Hence, Mallory made more shots then Christina!

--------------------------------------------------

Hope This Helped! Good Luck!

6 0

4 years ago

What is the GCF for 24, 48m

katrin2010 [14]

The GCF of 24 and 48 is 24.

48 / 24 = 2
24 / 24 = 1

4 0

3 years ago

Read 2 more answers

If ava makes $15 in 2 hours, what is her rate of pay?

zalisa [80]

Answer:

$7.50 per hour

Step-by-step explanation:

15/2=7.5

7 0

3 years ago

Read 2 more answers