Ajutatima la ecuatia q*435=810

One estimate of the population of the world on January 1, 2005, is

inn [45]

Answer:

8,566,379,470 people

Step-by-step explanation:

Let's start simple. In order to find the population increase on January 1, 2006, we need to multiply 6,486,915,022 by 1.4% and add it to 6,486,915,022.

6,486,915,022*1.4% = 90,816,810.308
90,816,810.308+6,486,915,022 = 6,577,731,832.31 people on January 2006.

Note that the above two steps gives the same answer as 6,486,915,022*1.014.

So we need to do this for each year. 20 years pass between 1/1/2005 and 1/1/2025.

We need to do 6,486,915,022*1.014*1.014*1.014... 20 times.

This is equivalent to $6,486,915,022*1.014^{20}$ .

Multiplying it out gives us 8566379470.2 = 8,566,379,470 people.

4 0

3 years ago

What is this equation? Solve for X-2/7=5/7

Anna [14]

x−2/7=5/7
so this means x = 1

5 0

4 years ago

Read 2 more answers

Lily bought 25.29 pounds of grapefruit. The lightest grapefruit weighed 1.4 pounds. The heaviest grapefruit weighed 1.6 pounds.

Gemiola [76]

Answer:

B

Step-by-step explanation:

GIven that the lightest is 1.4 and the heaviest is 1.6, this implies that ALL the fruit must weigh between 1.4 and 1.6.

This also implies that the average weight of the fruit must also be between 1.4 lb and 1.6 lb.

recall that : average weight = total weight / number of fruit

we are given than total weight = 25.29 lb, so

average weight = 25.29 / number of fruit

Just go down the choices and test which choice gives an average weight between 1.4 and 1.6 lbs.

Choice A: number of fruit = 19

average = 25.29 / 19 = 1.33 (outside of range, not the answer)

Choice B: number of fruit = 17

average = 25.29 / 17 =1.49 (within range , possible answer)

Choice C: number of fruit = 10

average = 25.29 / 10 = 2.529 (outside of range, not the answer)

Choice D: number of fruit = 50

average = 25.29 / 50 = 0.5058 (outside of range, not the answer)

From the above, we can see that only B gives an average between 1.4 and 1.6 lb.

5 0

3 years ago

Julie has basketball practice every 6 days and track practice every 4 days. In how many days will she have both basketball and t

GrogVix [38]

Answer: 12 days

Step-by-step explanation:

For us to know the number of days whereby Julie will have both basketball and track practice, we've to find the lowest common multiple of both 6 and 4. This will be:

Multiples of 4 = 4, 8, 12, 16, 24.

Multiples of 6 = 6, 12, 18, 24, 30.

The lowest common multiple is 12. Therefore, she'll have basketball and track practice in 12 days

5 0

3 years ago

A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on

SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± $Z_{1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

$Z_{1-\alpha /2}= Z_{0.95}= 1.645$

[10.41±1.645* $(\frac{3.71}{\sqrt{28} } )$ ]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0

4 years ago