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2 years ago

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Aaron borrows $150 from his friend Austin. He promises to pay back the money in 4 monthly installments. Each month he wants to p

ay half the amount he paid the previous month. Assuming Austin does not charge any interest, how much should Aaron pay the first month to repay the money as scheduled?
A. $60
B. $70
C. $80
D. $90
E. $100

1 answer:

Maurinko [17]2 years ago

5 0

Answer C. $80

80-40-20-10=150

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Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0

2 years ago

Match each polynomial expression to its additive inverse

zvonat [6]

Answer:

(1, 2, 3, 4) matches (A, C, B, D)

Step-by-step explanation:

The additive inverse of a term is formed by changing the sign. The additive inverse of the polynomial is formed by changing the sign of every term.

If we call the expressions on the left (top-to-bottom) 1, 2, 3, 4, and those on the right A, B, C, D, then the match-up <em>in this presentation of the question</em> is ...

1 - A

2 - C

3 - B

4 - D

_____

YMMV if the expressions are mixed differently

6 0

2 years ago

Read 2 more answers

According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg

Sati [7]

Answer:

a) Mean blood pressure for people in China.

b) 38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c) 71.30% probability that a person in China has blood pressure of 141 mmHg or less.

d) 8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e) Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg

f) 157.44mmHg

Step-by-step explanation:

When the distribution is normal, we use 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.

If X is two standard deviations from the mean or more, it is considered unusual.

In this question:

$\mu = 128, \sigma = 23$

a.) State the random variable.

Mean blood pressure for people in China.

b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.

This is 1 subtracted by the pvalue of Z when X = 135.

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

$Z = \frac{135 - 128}{23}$

$Z = 0.3$

$Z = 0.3$ has a pvalue of 0.6179

1 - 0.6179 = 0.3821

38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.

This is the pvalue of Z when X = 141.

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

$Z = \frac{141 - 128}{23}$

$Z = 0.565$

$Z = 0.565$ has a pvalue of 0.7140

71.30% probability that a person in China has blood pressure of 141 mmHg or less.

d.)Find the probability that a person in China has blood pressure between 120 and 125 mmHg.

This is the pvalue of Z when X = 125 subtracted by the pvalue of Z when X = 120. So

X = 125

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

$Z = \frac{125 - 128}{23}$

$Z = -0.13$

$Z = -0.13$ has a pvalue of 0.4483

X = 120

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

$Z = \frac{120 - 128}{23}$

$Z = -0.35$

$Z = -0.35$ has a pvalue of 0.3632

0.4483 - 0.3632 = 0.0851

8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?

From b), when X = 135, Z = 0.3

Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg.

f.) What blood pressure do 90% of all people in China have less than?

This is the 90th percentile, which is X when Z has a pvalue of 0.28. So X when Z = 1.28. Then

X = 120

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

$1.28 = \frac{X - 128}{23}$

$X - 128 = 1.28*23$

$X = 157.44$

So

157.44mmHg

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2 years ago

One coin is randomly selected from a jar containing 20 pennies, 15 nickels, 3 dimes, and 12 quarters. What is the odds of a valu

Aliun [14]

Total coins in the jar = 50. Number of coins with a value less than $0.05 = 20 (the pennies). ... Probability of picking a penny = 20/50 = 40%. ... 'Odds' = 3 to 2 against it.

6 0

2 years ago

In each represent the common form of each argument using letters to stand for component sentences, and fill in the blanks so tha

arlik [135]

Answer:

a

$a \to b$

$\neg b$

$\neg a$

b

If <u>all prime numbers are odd,</u> then <u>2 is odd</u>.

2 is not odd.

Therefore, it is not the case that all prime numbers are odd.

Step-by-step explanation:

Considering part A

The first sentence is

If all computer programs contain errors, then this program contains an error.

The second sentence is

This program does not contain an error.

The third sentence is

Therefore, it is not the case that all computer programs contain errors.

Now we will use a and b as the letter to denote the component sentences and $\neg a$ and $\neg b$ when the component sentences is not the case

So a = If all computer programs contain errors

and

b = this program contains an error.

Generally then is represented as $\to$

Hence the first sentence is denoted as

$a \to b$

The second sentence is represented as

$\neg b$

The third sentence is represented as

$\neg a$

So part a can be represented as

$a \to b$

$\neg b$

$\neg a$

Considering part B

Here the objective is to fill in the blank spaces so that the logic of the sentence in part b is as that in part a

Now comparing the second statement a and b

"This program does not contain an error" = "2 is not odd" = $\neg b$

Hence "this program contains an error." = "2 is odd" = b

Now comparing the third statement a and b

Therefore, it is not the case that all computer programs contain errors.

=

Therefore, it is not the case that all prime numbers are odd

=

$\neg a$

Hence

If all computer programs contain errors, = if all prime numbers are odd.= a

3 0

2 years ago