-14 = S-372 - -24<br> Need help solving this problem for math

The type of flower shown in the table make up an arrangement what percent of the flowers in the range that are roses

sdas [7]

Can you show the table please?

4 0

3 years ago

Show that the sum of two concave functions is concave. Is the product of two concave functions also concave?

spayn [35]

Answer with explanation:

Let us assume that the 2 functions are:

1) f(x)

2) g(x)

Now by definition of concave function we have the first derivative of the function should be strictly decreasing thus for the above 2 function we conclude that

$\frac{d}{dx}\cdot f(x)$

Now the sum of the 2 functions is shown below

$y=f(x)+g(x)$

Diffrentiating both sides with respect to 'x' we get

$\frac{dy}{dx}=\frac{d}{dx}\cdot f(x)+\frac{d}{dx}\cdot g(x)\\\\$

Since each term in the right of the above equation is negative thus we conclude that their sum is also negative thus

$\frac{dy}{dx}$

Thus the sum of the 2 functions is also a concave function.

Part 2)

The product of the 2 functions is shown below

$h=f(x)\cdot g(x)$

Diffrentiating both sides with respect to 'x' we get

$h'=\frac{d}{dx}\cdot (f(x)\cdot g(x))\\\\h'=g(x)f'(x)+f(x)g'(x)$

Now we can see the sign of the terms on the right hand side depend on the signs of the function's themselves hence we remain inconclusive about the sign of the product as a whole. Thus the product can be concave or convex.

8 0

3 years ago

The probability that your call to a service line is answered in less than 30 seconds is 0.85. Assume that your calls are indepen

aev [14]

Answer:

a) 0.1720

b) 0.8298

c) 19

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.85$

(a) If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X = 9)$ when $n = 12$ .

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{12,9}.(0.85)^{9}.(0.15)^{3} = 0.1720$

(b) If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X \geq 16)$ when $n = 20$