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padilas [110]
3 years ago
11

If Ann correctly factors an expression that is the difference of two perfect squares, her factors could be

Mathematics
1 answer:
Sphinxa [80]3 years ago
3 0

Answer:

1.) ???

2.) (2x+3y)(2x-3y)=4x²-9y²

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HEYYYYY YOU......I NEED HELP
Ivahew [28]

Answer:

3.50h+10

Step-by-step explanation:

H is hours

6 0
3 years ago
Read 2 more answers
I need help on 6, 7, and 8. Please help me fast!
Len [333]

Answer:

7. 130 + 150 = 280

360 - 280 = 80 degrees

Step-by-step explanation:

4 0
3 years ago
the population of Watesville decreases at a 1.6% per year.If the population was 62,500 in 2014 what will it be in 2020
vichka [17]

The population in 2020 will be 57657.5 approximately

<u>Explanation:</u>

Given:

Population in 2014, P = 62,500

Rate of decrease, r = 1.6%

Population in 2020, Q = ?

Number of years between 2014 to 2020 = 5 years

Population in 2020 = P(1 - \frac{r}{100} )^t

Substituting the value we get:

Q = 62500(1- \frac{1.6}{100})^5\\ \\Q = 62500 ( 0.92252) \\\\Q = 57657.46

Therefore, the population in 2020 will be 57657.5 approximately

7 0
2 years ago
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
Value times frequency equals product.<br> True or False
ElenaW [278]
True. Hope this helps!

5 0
3 years ago
Read 2 more answers
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