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Andrej [43]
3 years ago
12

which equation represents an exponential function that passes through the point (2, 80)? f(x) = 4(x)5 f(x) = 5(x)4 f(x) = 4(5)x

f(x) = 5(4)x
Mathematics
1 answer:
pav-90 [236]3 years ago
7 0

we know that

if the exponential function passes through the given point, then the point must satisfy the equation of the exponential function

we proceed to verify each case  if the point (2,80) satisfied the exponential function

<u>case A</u>  f(x)=4(x^{5})

For x=2 calculate the value of y in the equation and then compare with the y-coordinate of the point

so

f(2)=4(2^{5})=128

128\neq 80

therefore

the exponential function  f(x)=4(x^{5}) not passes through the point (2,80)

<u>case B</u>  f(x)=5(x^{4})

For x=2 calculate the value of y in the equation and then compare with the y-coordinate of the point

so

f(2)=5(2^{4})=80

80=80

therefore

the exponential function f(x)=5(x^{4}) passes through the point (2,80)

<u>case C</u>  f(x)=4(5^{x})

For x=2 calculate the value of y in the equation and then compare with the y-coordinate of the point

so

f(2)=4(5^{2})=100

100\neq 80

therefore

the exponential function f(x)=4(5^{x}) not passes through the point (2,80)

<u>case D</u>  f(x)=5(4^{x})

For x=2 calculate the value of y in the equation and then compare with the y-coordinate of the point

so

f(2)=5(4^{2})=80

80=80

therefore

the exponential function f(x)=5(4^{x}) passes through the point (2,80)

therefore

<u>the answer is</u>

f(x)=5(x^{4})

f(x)=5(4^{x})

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An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos
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Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.78, n = 675

So

\mu = E(X) = np = 675*0.78 = 526.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when X = 0.8*675 = 540. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{540 - 526.5}{10.76}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

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