250 is what percent of 500

NEED THE ANSWER ASAP PLEASE

PolarNik [594]

Answer:

lol

Step-by-step explanation:

xd

6 0

3 years ago

6 dogs are black, 5 dogs are white, and the rest are blue. If there are 15 dogs in total, how many are blue?

Bezzdna [24]

Answer:

4 blue dogs

Step-by-step explanation:

Since there are 6 black dogs and 5 white dogs, there is 11 black and white dogs. There is 15 dogs in total, so you subtract 11 from 15, and that is 4. So, 4 dogs are blue.

6 0

3 years ago

A soccer league has 160 players. Of those players 40% are boys. How many boys are in the soccer league?

MatroZZZ [7]

The answer is 64 boys are in the soccer league.

X/160 = 40/100

X/160 * 100 = 40/100 * 100

5x/8 = 40

(8 * 5x)/8 = 40 * 8

5x = 320

5x/5 = 320/5

x = 64

7 0

3 years ago

There are 15 identical pens in your drawer, nine of which have never been used. On Monday, yourandomly choose 3 pens to take wit

DaniilM [7]

Answer: p = 0.9337

Step-by-step explanation: from the question, we have that

total number of pen (n)= 15

number of pen that has never been used=9

number of pen that has been used = 15 - 9 =6

number of pen choosing on monday = 3

total number of pen choosing on tuesday=3

note that the total number of pen is constant (15) since he returned the pen back .

probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

probability of picking a pen that has been used on tuesday = 6/15 = 2/5

probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5

on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .

we know that

probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.

to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution

p(x=r) = $nCr * p^{r} * q^{n-r}$

n= total number of pens=15

r = number of pen chosen on tuesday = 3

p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

by slotting in the parameters, we have that

p(x=3) = $15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}$

p(x=3) = 455 * $0.4^{3} * 0.6^{12}$

p(x=3) = 455 * 0.064 * 0.002176

p(x=3) = 0.0633

thus probability that 3 of the pens has been used on tuesday. = 0.0633

probability of ball that none of the 3 pen has never being used on tuesday = 1 - 0.0633 = 0.9337

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.

6 0

3 years ago