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3 years ago

There were 935 offenses in 2000 and 2448 offenses in 2019. What is the percentage increase?

Mathematics

1 answer:

Artist 52 [7]3 years ago

6 0

Answer:

Approximately 262%

Step-by-step explanation:

To find the percentage increase divide the new amount of offenses by the old amount of offenses:

2448/935 = 2.618

2.618 rounds to 2.62. Convert 2.62 into a percentage: 262%

The percentage increase between the amount of offenses in 2000 and the amount of offenses in 2019 is approximately 262%.

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What is the simplest form of 2/10?

bezimeni [28]

<span>To get to the simplest form you have to divide numerator and denominator by numbers they can both be divided by, until you get to a fraction where you can't divide both anymore getting an integer as a result.
Here you can only divide by 2 and you get 1/5. That's the simplest form</span>, if you kept dividing you'd get decimals, but the simplest form must be integers.

4 0

3 years ago

Solving Rational equations. LCD method. Show work. Image attached.

Ugo [173]

Answer:

$k=\frac{1}{2}$

Step-by-step explanation:

The given rational equation is

$\frac{5k}{k+2}+\frac{2}{k}=5$

The Least Common Denominator is

$k(k+2)$

Multiply each term by the LCD.

$k(k+2)\times \frac{5k}{k+2}+k(k+2)\times \frac{2}{k}=5k(k+2)$

Simplify;

$k\times \frac{5k}{1}+(k+2)\times \frac{2}{1}=5k(k+2)$

$\Rightarrow k(5k)+2(k+2)=5k(k+2)$

Expand;

$\Rightarrow 5k^2+2k+4=5k^2+10k$

Group similar terms;

$\Rightarrow 5k^2-5k^2+2k-10k=-4$

$\Rightarrow -8k=-4$

Divide by -8.

$k=\frac{1}{2}$

5 0

3 years ago

Suppose the parent population has an exponential distribution with a mean of 15 and standard deviation of 12. Use the Central Li

bazaltina [42]

Answer:

The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the population, we have that:

Mean = 15

Standard deviaiton = 12

Sample of 30

By the Central Limit Theorem

Mean 15

Standard deviation $s = \frac{12}{\sqrt{30}} = 2.19$

Approximately normal

The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.

4 0

3 years ago

PLEASE HELP ME ON (A)

Liula [17]

It’s 4 ik know this it’s four cixhdhdud didjdhdh djdjdhd djdjdhd

3 0

3 years ago

Read 2 more answers

An airplane must clear a 60-foot pole at the end of a runway 500 yards long determine the angle of elevation at which the airpla

Marat540 [252]

Answer:

2.3 degrees.

Step-by-step explanation:

Please find the attachment.

We are told that an airplane must clear a 60-foot pole at the end of a runway 500 yards long.

Let us convert 500 yards to feet.

1 yard= 3 feet.

500 yards= 3*500 feet= 1500 feet.

We can see from our attachment pole and runway are in form of a right triangle. The pole is opposite to angle of elevation of plane and length of runway is adjacent.

Since tangent represents the relation between opposite and adjacent of right triangle, So we will use tangent to find angle of elevation that plane must ascend to clear the pole.

$tan(\theta)=\frac{60}{1500}$

$\theta=\tan^{-1}(\frac{60}{1500} )$

$\theta=\tan^{-1}(0.04)$

$\theta=2.290610042639$

Therefore, the airplane must ascend 2.3 degrees to clear the pole.

8 0

3 years ago