What caused the Mongols to lose the battle at Ain Jalut? *

In 2010, the population of Kingsford was 8000. By 2014 the population had increased by 15% and 2/5 of the population was age 12

Anastasy [175]

<h2>Hello!</h2>

The answer is: The population age 12 or under was 3680.

If in 2010 the population of Kignsford was 8000, and by 2014 the population had increased by 15% we need to calculate the total population and then, calculate how many people was age 12 or under.

We must remember that we cannot work with percentage values, so, to make 15% a real number, we need to divide it by 100:

$15(Percentage)=\frac{15}{100}=0.15$

So,

Calculating the population by 2014, we have:

$Population_{2014}=Population_{2010}+ Population_{2010}*0.15$

Then, substituting we have:

$Population_{2014}=8000+ 8000*0.15=9200$

Now, calculating the population that was age 12 or under, knowing that it was 2/5 of the total population, we have:

$Population=TotalPopulation*\frac{2}{5}\\\\Population=9200*\frac{2}{5}=3680$

So, the population age 12 or under was 3680.

Have a nice day!

6 0

3 years ago

Identify one unique decimal that you encounter regularly during the week and how it affects you or not in your daily life. your

Tpy6a [65]

A unique decimal that I encounter regularly is "0.05". I encounter this almost every day, when I go to the store to buy something, the saleslady gives me $0.05 as change. When I go home and put some of my extra money in my piggy bank, I hold a $0.05 in my hand.

5 0

2 years ago

Salaries of 49 college graduates who took a statistics course in college have a mean, x overbar, of $ 65 comma 300. Assuming a

Molodets [167]

Answer: $60,540< \mu$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}-z^*\dfrac{\sigma}{\sqrt{n}}< \mu$

, where $\sigma$ = Population standard deviation.

n= sample size

$\overline{x}$ = Sample mean

z* = Critical z-value .

Given : $\sigma=\$17,000$

n= 49

$\overline{x}= \$65,300$

Two-tailed critical value for 95% confidence interval = $z^*=1.960$

Then, the 95% confidence interval would be :-

$65,300-(1.96)\dfrac{17000}{\sqrt{49}}< \mu$

$=65,300-(1.96)\dfrac{17000}{7}< \mu$

$=65,300-4760< \mu$

$=60,540< \mu$

Hence, the 95% confidence interval for estimating the population mean $(\mu)$ :

$60,540< \mu$

7 0

2 years ago

Celina measured the height of a window frame as 5.55 feet, but the actual height was 6 feet. What is the percentage of error in

REY [17]

Is this Dolli Buchannan Its Alexis!!!!

8 0

2 years ago

A bank loaned out $28,500, part of it at the rate of 14% annual interest, and the rest at 4% annual

Flauer [41]

Answer:

$15,000 was invested at 14% per year, and $13,500 was invested at 4% per year.

Step-by-step explanation:

Given that a bank loaned out $ 28,500, part of it at the rate of 14% annual interest, and the rest at 4% annual interest, and the total interest earned for both loans was $ 2,640.00, to determine how much was loaned at each rate you must perform the following calculations:

28,500 x 0.14 + 0 x 0.04 = 3,990

20,000 x 0.14 + 8,500 x 0.04 = 3,140

15,000 x 0.14 + 13,500 x 0.04 = 2,640

Therefore, $ 15,000 was invested at 14% per year, and $ 13,500 was invested at 4% per year.

5 0

2 years ago