Find the quotient:<br> 3)21<br> A. 6<br> B. 7<br> C. 8<br> D. 9

Im stuck on this problem. Please explain how you got your answer

klemol [59]

The function $y=0.3^x$ represents exponential growth with the initial value equal to 1, the decay factor equal to 0.3, and the rate equal to 0.7.

<h3>Population Growth Equation</h3>

The formula for the Population Growth Equation is:

$P_f=P_o*(1+\frac{R}{100}) ^t$

Pf= future population

Po=initial population

r=growth rate

t= time (years)

growth or decay factor = (1 ±r)

When 1+R > 1, the equation represents growth, while 1+R < 1 the equation represents decay.

The question gives:

$y=0.3^x$ , then

Pf=y

Po= 1

$(1+\frac{R}{100}) =0.3$ , thus

$(100+R}) =30\\ \\ R=-100+30\\ \\ R=-70$

r= -70%= -0.7

decay factor= (1-0.7)=0.3

Therefore,

1+R will be = 1+(-0.7)=1 - 0.7 =0.3

When 1+R >1, the function represents exponential growth.

Read more about the exponential function here:

brainly.com/question/8935549

4 0

2 years ago

What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success p? C

charle [14.2K]

Answer:

$(D)E[ X ] =np.$

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success p,

$f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq x\leq n$

$E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}$

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

$E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}$

Now,

$x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)$