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3 years ago

Determine all critical points for the function. f(x)=(x-10)^5

1 answer:

7 0

A critical point is just where a graph's slope is = to 0, so you can take the derivative of that function and set it equal to 0.

Using the chain rule:
f'(x) = 5(x-10)^4
0 = 5(x-10)^4

divide by 5 on both sides and take the 4th root of both sides.

0 = x - 10

add ten on both sides, and x = 10. Now, plug the x-value back into the original (you want to find the y-value on the original function ,not the derivative).

y = (10-10)^5 = 0

(10,0)

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Match the verbal expression (term) with its algebraic expression (definition).

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

A candy company claims that 20% of the candies in its bags are colored green. Steve buys 30 bags of 30 candies, randomly selects

Allisa [31]

Answer:

The probability of Steve agreeing with the company’s claim is 0.50502.

Step-by-step explanation:

Let X denote the number of green candies.

The probability of green candies is, p = 0.20.

Steve buys 30 bags of 30 candies, randomly selects one candy from each, and counts the number of green candies.

So, n = 30 candies are randomly selected.

All the candies are independent of each other.

The random variable X follows a binomial distribution with parameter n = 30 and p = 0.20.

It is provided that if there are 5, 6, or 7 green candies, Steve will conclude that the company’s claim is correct.

Compute the probability of 5, 6 and 7 green candies as follows:

$P(X=5)={30\choose 5}(0.20)^{5}(1-0.20)^{30-5}=0.17228\\\\P(X=6)={30\choose 6}(0.20)^{6}(1-0.20)^{30-6}=0.17946\\\\P(X=7)={30\choose 7}(0.20)^{7}(1-0.20)^{30-7}=0.15328$

Then the probability of Steve agreeing with the company’s claim is:

P (Accepting the claim) = P (X = 5) + P (X = 6) + P (X = 7)

= 0.17228 + 0.17946 + 0.15328

= 0.50502

Thus, the probability of Steve agreeing with the company’s claim is 0.50502.

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3 years ago

Factories A and B produce computers. Factory A produces 3 times as many computers as factory B. The probability that an item pro

OleMash [197]

Answer:

P(A∣D) = 0.667

Step-by-step explanation:

We are given;

P(A) = 3P(B)

P(D|A) = 0.03

P(D|B) = 0.045

Now, we want to find P(A∣D) which is the posterior probability that a computer comes from factory A when given that it is defective.

Using Bayes' Rule and Law of Total Probability, we will get;

P(A∣D) = [P(A) * P(D|A)]/[(P(A) * P(D|A)) + (P(B) * P(D|B))]

Plugging in the relevant values, we have;

P(A∣D) = [3P(B) * 0.03]/[(3P(B) * 0.03) + (P(B) * 0.045)]

P(A∣D) = [P(B)/P(B)] [0.09]/[0.09 + 0.045]

P(B) will cancel out to give;

P(A∣D) = 0.09/0.135

P(A∣D) = 0.667

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3 years ago

If s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n st ?

yulyashka [42]

Answer:

A) None

Step-by-step explanation:

1) $s^t$ shoudnt neccesarily be a factor of nst, for example, if s = 3, t = 4, and n = 12, then both s and t are factors of n, but $3^4 = 81$ is not a factor of nst = 144.

2) $(st)^2$ shoudnt neccesarily be a factor of nst. Let s be 4, let t be 6, and let n be 12. Then n is a factor of both s and t, but $(st)^2 = 24^2$ is not a factor of nst = 12*24. In fact, it is a greater number.

3) Again, s+t isnt necessarily a factor of nst, let s be 2 and t be 3. Then both s and t are factor of n = 12. However 5 = s+t is not a factor of nst = 72.

So, neither of the three options is guaranteed to be a factor of nst. In fact, for s = 4, t = 6, and n = 12, none of the three options are valid.

4 0

3 years ago