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lianna [129]
2 years ago
15

A function, f(x)=4x+5, has a domain 0<=x<=50. What is the range? How do you do this??

Mathematics
1 answer:
babunello [35]2 years ago
3 0

Step-by-step explanation:

remember the domain is the interval or set of valid input (x) values. the range is the interval or set of the valid result (y) values.

so, the given domain tells us the interval to look at.

what are the functional values of the function between 0 and 50 ?

since the function is continuous (there are no gaps) in this interval, we can safely assume that all values between f(0) and f(50) are valid y values.

f(0) = 4×0 + 5 = 5

f(50) = 4×50 + 5 = 205

so, the range is 5 <= y <= 205

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Answer:

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Step-by-step explanation:

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3 years ago
Not all visitors to a certain company's website are customers. In fact, the website administrator estimates that about 5% of all
Gnom [1K]

Answer:

0.0135 = 1.35% probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

Step-by-step explanation:

For each visitor of the website, there are only two possible outcomes. Either they are looking for the website, or they are not. The probability of a customer being looking for the website is independent of other customers. This means that the binomial probability distribution is used 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.

5% of all visitors to the website are looking for other websites.

So 100 - 5 = 95% are looking for the website, which means that p = 0.95

Find the probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

This is P(X = 2) when n = 4. So

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

P(X = x) = C_{4,2}.(0.95)^{2}.(0.05)^{2} = 0.0135

0.0135 = 1.35% probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

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2 years ago
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7 0
2 years ago
Suppose that your business is operating at the 4.5-Sigma quality level. If projects have an average improvement rate of 50% annu
Luda [366]

Answer:

  11.75 years

Step-by-step explanation:

If we ignore the fact that "6-sigma" quality means the error rate corresponds to about -4.5σ (3.4 ppm) and simply go with ...

  P(z ≤ -6) ≈ 9.86588×10^-10

and

  P(z ≤ -4.5) ≈ 3.39767×10^-6

the ratio of these error rates is about 0.000290372. We're multiplying the error rate by 0.5 each year, so we want to find the power of 0.50 that gives this value:

  0.50^t = 0.000290372

  t·log(0.50) = log(0.00290372) . . . . take logarithms

  t = log(0.000290372)/log(0.50) ≈ -3.537045/-0.301030

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It will take about 11.75 years to achieve Six Sigma quality (0.99 ppb error rate).

_____

<em>Comment on Six Sigma</em>

A 3.4 ppm error rate is customarily associated with "Six Sigma" quality. It assumes that the process may have an offset from the mean of up to 1.5 sigma, so the "six sigma" error rate is P(z ≤ (1.5 -6)) = P(z ≤ -4.5) ≈ 3.4·10^-6.

Using that same criteria for the "4.5-Sigma" quality level, we find that error rate to be P(z ≤ (1.5 -4.5)) = P(z ≤ -3) ≈ 1.35·10^-3.

Then the improvement ratio needs to be only 0.00251699, and it will take only about ...

  t ≈ log(0.00251699)/log(0.5) ≈ 8.6 . . . . years

5 0
3 years ago
Please help I need it please please
Anni [7]

Answer:

23 %

Step-by-step explanation:

151.34 ÷ 658 × 100

hope this helps

have a nice day <3

6 0
2 years ago
Read 2 more answers
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