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3 years ago

a country's population in 1992 waa 83 million and in 1996 was 88 million what was the population in 2013

1 answer:

4 0

The population in 1992 was 83 mill....population is 1996 was 88 mill...

so in 4 years (from 1992 to 1996), the population rose by (88 mill - 83 mill) = 5 mill

Thats 5,000,000 / 4 = 1,250,000 per year.

so in 2013 (17 years from 1996), the population would be :
83 mill + 1,250,000(17) = 83 mill + 2,125,000 = 85,125,000 <==

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Please please please help me I'm really confused and I don't understand it

maxonik [38]

Answer : p=2

2p-2-5x=-2(7-2x)

First, work on the parentheses. So the -2 on the outside as to be multiplied by the 7 , which gets -14 . Then you multiply the -2 by the 2x, which is 4x.

It will go from -2(7-2x) to -14+4x
But if the -14+4x looks too confusing, you can just switch them around to look “normal”. So it would be 4x-14 .

Now that we’re finished with one side of the problem, we’ll go to the other side.

On the left side , there are two variables, or letters. Because they are the same letter (p) all you have to do is add them together.

3p
-5p
+—-
-2p

And then just carry the -2 that was originally there , so it’s -2p-2

Now the equation is
-2p-2 = 4p-14

So we’re going to add 2p to both sides.

The -2p on the left cancels out, and the 4p on the right side becomes 6p

-2 = 6p-14

Next, you are going to want to isolate the variable. (Leave the letter by itself) so you add 14 to the -14 , so it can cancel out. (AND DO THE SAME ON THE OTHER SIDE)
So, you add 14 to the -2 on the left of the equation. Which comes out to 12.

12 = 6p

Lastly, divide the 6p , so you can get the letter on it’s own. And do the same on the other side.

12 divided by 6 is 2

2 = p

OR

p = 2

4 0

4 years ago

Whats the LCM of (50, 75)

True [87]

Answer:

What is the LCM of 50 and 75? The lcm of 50 and 75 is 150.

4 0

3 years ago

Read 2 more answers

Help please. Y’all are the best!!

Alchen [17]

The answer is D) -2

Parallel lines have the same slope. Parallel lines will just intersect the x and y axis in different spots.

Hope this helps!! :)

4 0

4 years ago

An airplane heads in the direction of N25°E at 325 mph. A tailwind is blowing in the direction N40°W at 50 mph. What is the actu

geniusboy [140]

Answer: I Think It’s A 315.786

Step-by-step explanation: Let me know if that’s correct

6 0

3 years ago

The average length of a field goal in the National Football League is 38.4 yards, and the s. d. is 5.4 yards. Suppose a typical

Tasya [4]

Answer:

a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

d-a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

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.

In this problem, we have that:

$\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538$

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

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

By the Central Limit Theorem

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

$Z = \frac{36 - 38.4}{0.8538}$

$Z = -2.81$

$Z = -2.81$ has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

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

$Z = \frac{41 - 38.4}{0.8538}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now $n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08$

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

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

$Z = \frac{36 - 38.4}{1.08}$