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1 year ago

812.16 is what percent of 752?

Mathematics

2 answers:

belka [17]1 year ago

8 0

Answer:

108

Step-by-step explanation:

For 100, our answer will be 108

For 752, our answer will be x

100*x = 108*752 (In Step 1 and 2 see colored text; Diagonal multiplications will always be equal)

x = 108*752/100 = 81216/100 = 812.16

Answer: 108 percent of 752 is 812.16

finlep [7]1 year ago

3 0

108 is the answer to your question

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Answer:

$A(t)=975(1.025)^t$

In 2025,the number of students at the villages high school=1159

Step-by-step explanation:

We are given that in 2018

Number of students at the villages high school=975

Increasing rate,r=2.5%=0.025

We have to write and use of exponential growth function to project the populating in 2025.

$A_0=975,t=0$

According to question

Number of students at the villages high School is given by

$A(t)=A_0(1+r)^t$

Substitute the values

$A(t)=975(1+0.025)^t=975(1.025)^t$

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Step-by-step explanation:

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2 years ago

The ratio of 3 km to 1800 m is

Ugo [173]

5 : 3

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2 years ago

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According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

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Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

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