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2 years ago

A basketball player made 80 out of 100 attempted free throws what percent of free throws was made ?

Mathematics

2 answers:

AveGali [126]2 years ago

8 0

Answer:

80%

Step-by-step explanation:

80÷100=.8 which is 80%

Tom [10]2 years ago

6 0

Answer:

8/10

Step-by-step explanation:

id hope your have a great day

You might be interested in

Are the following expressions equivalent 26-(-26) and 26+(-26)

kotegsom [21]

Answer:

No.

Step-by-step explanation:

They are not because the left side is 52, while the right side is 0.

5 0

3 years ago

A random sample has been taken from a normal distribution and the following confidence intervals constructed using the same data

LuckyWell [14K]

Answer:

(25.53, 37.87): 95% CI

(23.59, 39.81): 99% CI

Step-by-step explanation:

The margin of error of a confidence interval is given by:

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

z is related to the confidence level. The higher the confidence level, the higher the values of z, and thus, we wider the confidence interval is.

In this question:

The narrower C.I. is the 95%, and the wider is the 99%. So

(25.53, 37.87): 95% CI

(23.59, 39.81): 99% CI

4 0

2 years ago

Camille had $275 in her checking account . she deposited $812 and wrote checks $75, $60, and $48. how much did she have in the a

Elenna [48]

She would have $904. 275 + 812= 1087. 75 + 60 + 48= 183. 1087 - 183 = 904. Check the math to make sure, I did it in my head.

3 0

3 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.