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3 years ago

Why do we study statistics?

Mathematics

1 answer:

8_murik_8 [283]3 years ago

3 0

Answer:

B. Because statistics can be applied to every field of study

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WILL MARK BRAINLIEST<br><br>Which equation can be used to find the volume of this solid?<br>

AURORKA [14]

V=(7*6)/2 *8 the third option

you multiply the height with the base of the triangle and you get the area of the base of the prism so you multiply it with the height of the prism to get the volume

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3 years ago

Twenty minutes to six in the evening on a 24 hour clock

Taya2010 [7]

Answer:

1740 is the time on a 24 hour clock or military time

6 0

3 years ago

For what values of m does the graph of y = 3x^2+ 7x + m have two x-intercepts?

qaws [65]

Answer:

m < 49/12

Step-by-step explanation:

The portion of the quadratic formula under the square root sign is the discriminant.

If the discriminant is > 0 then there are two real roots.

b² -4ac > 0

-----------------------------

7² - 4(3)m > 0

49 - 12m > 0

Subtract 49 from both sides

-12m > -49

Divide both sides by -12

(when multiplying or dividing by a negative the inequality must be reversed)

m < 49/12

8 0

3 years ago

Round to the nearest ten for 2,420

kotykmax [81]

I'm pretty sure it's just 2,420.

5 0

3 years ago

Read 2 more answers

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

GaryK [48]

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:

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

3 0

3 years ago