The cast of the play makes 275 programs for 5 shows. How many programs will be needed for 13 shows?

how many 4 digit numbers n have the property that the 3 digit number obtained by removing the leftmost digit is one ninth of n

Tom [10]

Answer:

7

Step-by-step explanation:

We want to find the number 4-digit of positive integers n such that removing the thousands digit divides the number by 9.

__

Let the thousands digit be 'd'. Then we want to find the integer solutions to ...

n -1000d = n/9

n -n/9 = 1000d . . . . . . add 1000d -n/9

8n = 9000d . . . . . . . . multiply by 9

n = 1125d . . . . . . . . . divide by 8

The values of d that will give a suitable 4-digit value of n are 1 through 7.

When d=8, n is 9000. Removing the 9 gives 0, not 1000.

When d=9, n is 10125, not a 4-digit number.

There are 7 4-digit numbers such that removing the thousands digit gives 1/9 of the number.

8 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

Mamont248 [21]

Answer:

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

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

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

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 50, \sigma = 6$

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

This is 1 subtracted by the pvalue of Z when X = 60.

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

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

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

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

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

This is the pvalue of Z when X = 60.

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

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

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

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

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

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

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

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

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

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

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

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

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

This is 1 subtracted by the pvalue of Z when X = 76.

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

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

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

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

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

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

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

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

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

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

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

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

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

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

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0

3 years ago

4x2 + 4x =- 4 Find the solutions to the Quadratic equation.

VashaNatasha [74]

Answer:

x=-3

Step-by-step explanation:

(4)(2) + 4x = -4

multiply the 4 and 2

8 + 4x = -4

subtract 8 from both sides

4x = -12

divide both sides by 4

x = -3

7 0

3 years ago

5/6 - 1/2 = ? A. 1/4 B. 1/3 C. 3/4 D. 2/3

zheka24 [161]

5. . 1
---- - -----
6. . 2

5. . 3
----- - -----
6. . 6

5-3
-------
6

2
-----
6

= 1/3

B

3 0

3 years ago

Georgia’s gross pay was $35,600 this year. She is to pay a federal income tax of 16%. How much should Georgia pay in federal inc

LenaWriter [7]

Georgia should pay $5,260.26 federal tax.

I am assuming that Georgia is an employee.

Gross Pay 35,600.00
Deductions:
C.O.L.A
Social Security 6.20% 2,207.20
Medicare 1.45% 516.20 2,723.40 
Taxable Income 32,876.60
Federal Income Tax 16% 5,260.26
Net Income 27,616.34

Gross income is different from Taxable Income. There are deductions that must be deducted from gross income to arrive at the taxable income. Deductions such as: Cost of Living Allowances (Social Security, Medicare, Retirement Earnings, other non taxable deductions).

There are also deductions that are subject to tax but not the federal estate tax. They are deductible because the tax for that item has already been paid in another form of tax. If said item is not deducted, then double taxation occurs, meaning the taxpayer paid two kinds of taxes for only one item.

8 0

3 years ago

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