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Aleonysh [2.5K]

2 years ago

6

A small town's population, P, can be modeled by the equation P (t) = 45,000(2.66), where t is the time in years. Is the statemen

t below true of false. The rate of population growth each year is 166% O True O False

1 answer:

ICE Princess25 [194]2 years ago

3 0

A function assigns the values. The statement "The rate of population growth each year is 166%" is true.

<h3>What is a Function?</h3>

A function assigns the value of each element of one set to the other specific element of another set.

Given the function of the population for the town is $P (t) = 45,000(2.66)^t$ , where t is the time in years. Comparing the given function to the general function of the population,

P(x) = a(1+b)ˣ

Then, the value of b will be 166%.

Hence, the statement "The rate of population growth each year is 166%" is true.

Learn more about Function:

brainly.com/question/5245372

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If 7x - 4y = 23 and x + y = 8, what is the value of x?

Margaret [11]

Answer:

x=5

Step-by-step explanation:

7x - 4y = 23 and x + y = 8

Multiply the second equation by 4

4x + 4y = 32

Add this to the first equation

7x - 4y = 23

4x + 4y = 32

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

11x = 55

Divide each side by 11

11x/11 = 55/11

x = 5

8 0

3 years ago

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100 meters, 9.69 seconds, usain bolt jamaica

uranmaximum [27]

10.31991744 meters per second

6 0

3 years ago

What is the value of the expression “two less than six times the difference of a number and five” when n = 9?

Finger [1]

Answer:

Try solving 6(n-5)-2.

Step-by-step explanation:

I believe this can be set up as 6(n-5)-2. The difference of a number and five can be represented by n-5. If this needs to be multiplied by 6 (hence 6 times), it would become 6(n-5). Two less than that is represented by -2. Thus, two less than six times the difference of a number and five is 6(n-5) -2, which, when n=9 is plugged in, looks like 6(9-5) -2. Follow PEMDAS from there.

7 0

3 years ago

The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c

Anestetic [448]

Answer:

a) The mean is $\mu = 60$

b) The standard deviation is $\sigma = 9$

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions 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:

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

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.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when $X = 55.5, Z = -0.5$

So

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

$-0.5 = \frac{55.5 - \mu}{\sigma}$

$-0.5\sigma = 55.5 - \mu$

$\mu = 55.5 + 0.5\sigma$

The probability a student selected at random takes no more than 71.52 minutes to complete the examination equals 0.8997.

This means that when X = 71.52, Z has a pvalue of 0.8997. This means that when $X = 71.52, Z = 1.28$

So

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

$1.28 = \frac{71.52 - \mu}{\sigma}$

$1.28\sigma = 71.52 - \mu$

$\mu = 71.52 - 1.28\sigma$

Since we also have that $\mu = 55.5 + 0.5\sigma$

$55.5 + 0.5\sigma = 71.52 - 1.28\sigma$

$1.78\sigma = 71.52 - 55.5$

$\sigma = \frac{(71.52 - 55.5)}{1.78}$

$\sigma = 9$

$\mu = 55.5 + 0.5\sigma = 55.5 + 0.5*9 = 55.5 + 4.5 = 60$

Question

The mean is $\mu = 60$

The standard deviation is $\sigma = 9$

6 0

3 years ago

For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course

pav-90 [236]

By definition of absolute value, you have

$f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1$

or more simply,

$f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x$

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

$f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x$

Note the strict inequalities in the definition of <em>f '(x)</em>.

In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:

$\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1$

$\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1$

All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

4 0

3 years ago