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2 years ago

The function f has the property that, for all x, 5(f)x=f(5x). If f(20)=40, what is the value of f(4)?

Mathematics

1 answer:

Romashka-Z-Leto [24]2 years ago

3 0

Using the given property, we conclude that:

f(4) = 8

<h3>What is the value of f(4)?</h3>

We know that our function has the property:

5*f(x) = f(5*x)

Now, we also know that:

f(20) = 40

Notice that we can rewrite:

f(20) = f(5*4)

Using the above property, we know that:

f(20) = f(5*4) = 5*f(4) = 40

Using the last part:

5*f(4) = 40

f(4) = 40/5 = 8

f(4) = 8

If you want to learn more about functions:

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Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

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Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

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What is the slope of the line that passes through the points (-9, -8)((−15,−16)? Write your answer in simplest form.

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Answer:

Therefore the slope of the line that passes through the points (-9, -8),(−15,−16) is

$Slope=\dfrac{4}{3}$

Step-by-step explanation:

Given:

Let,

point A( x₁ , y₁) ≡ ( -9 ,-8)

point B( x₂ , y₂ )≡ (-15 ,-16)

To Find:

Slope = ?

Solution:

Slope of Line Segment AB is given as

$Slope(AB)=\dfrac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Substituting the values we get

$Slope(AB)=\dfrac{-16-(-8)}{-15-(-9)}\\\\Slope(AB)=\dfrac{-16+8}{-15+9}\\\\Slope(AB)=\dfrac{-8}{-6}=\dfrac{4}{3}$

Therefore the slope of the line that passes through the points (-9, -8),(−15,−16) is

$Slope(AB)=\dfrac{4}{3}$

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