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3 years ago

The population of a city increases by 5% every year. If its population today is 300,000, what will its population be

Mathematics

2 answers:

alexandr1967 [171]3 years ago

5 0

The Answer is : is that the population today is 330,750

Julli [10]3 years ago

4 0

Answer:

330,750 people

Step-by-step explanation:

If the population increases by 5% every year, in 2 years the population of the city will be 330,750 people. D is also the correct answer as shown in the file.

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Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

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

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

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

$Z = \frac{6262 - 5656}{333.3}$

$Z = 1.82$

$Z = 1.81$ has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

$P = 100% - 96.562% = 3.438%$ of the light bulbs will last more than 6262 hours.

(b) What proportion of light bulbs will last 5252 hours or less?

This is the pvalue of the zscore of $X = 5252$

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

$Z = \frac{5252- 5656}{333.3}$

$Z = -1.21$

$Z = -1.21$ has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 hours?

This is the pvalue of the zscore of $X = 6262$ subtracted by the pvalue of the zscore $X = 5858$

For $X = 6262$ , we have that $Z = 1.81$ with a pvalue of .96562.

For X = 5858

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

$Z = \frac{5858- 5656}{333.3}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

$P = .96562 - .72907 = .23655$

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

(d) What is the probability that a randomly selected light bulb lasts less than 4646 hours?

This is the pvalue of the zscore of $X = 4646$

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

$Z = \frac{4646- 5656}{333.3}$

$Z = -3.03$

$Z = -3.03$ has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0

3 years ago

What is the solution to3/2 log 9/125

REY [17]

Answer:

-1.714001255

Step-by-step explanation:

Well just plug it in your calculator.

$\frac{3}{2} log(\frac{9}{125} )$

4 0

2 years ago

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by ℎ = −16푡 ଶ+103푡+5 w

Liono4ka [1.6K]

Question:

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by h(t)=-16t^2+80t+5, where t is measured in seconds. Write an equation to determine how long it will take for the ball to reach the ground.

Answer:

$t = 5.0625$

Step-by-step explanation:

Given

$h(t)=-16t^2+80t+5$

Required

Find t when the ball hits the ground

This implies that h(t) = 0

So, we have:

$0=-16t^2+80t+5$

Reorder

$-16t^2+80t+5 = 0$

Using quadratic formula, we have:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = -16$ $b =80$ $c = 5$

So, we have:

$t = \frac{-80\±\sqrt{80^2 - 4*-16*5}}{2*-16}$

$t = \frac{-80\±\sqrt{6400 +320}}{-32}$

$t = \frac{-80\±\sqrt{6720}}{-32}$

$t = \frac{-80\±82.0}{-32}$

This gives:

$t = \frac{-80+82.0}{-32}$ or $t = \frac{-80-82.0}{-32}$

$t = \frac{2}{-32}$ or $t = \frac{-162}{-32}$

$t = -\frac{2}{32}$ or $t = \frac{162}{32}$

But time can not be negative.

So, we have:

$t = \frac{162}{32}$

$t = 5.0625$

<em>Hence, time to hit the ground is 5.0625 seconds</em>

3 0

3 years ago

Please help me in solving this 4Log216 = 12x

bezimeni [28]

Answer:

log(6) = x

Step-by-step explanation:

4log(216)=12x

Divide both sides by 12

log(216)/3 = x

Use properties of logs

log(216^(1/3)) = x

Simplify

log(6) = x

6 0

3 years ago

Read 2 more answers

Multiplying a number by 4/5 then dividing by 2/5 is the same as multiplying by what number?

Verizon [17]

Would it be 33/10? If you are multiplying and then division you would use the inverse of the divided fraction and multiply. So you could invert 2/5 to 5/2 and add to 4/5. After finding the common denominator it would be adding 8/10 + 25/10 = 33/10.

5 0

3 years ago