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3 years ago

Evaluate the function f(x) at the given numbers (correct to six decimal places).

Mathematics

1 answer:

statuscvo [17]3 years ago

8 0

Answer:

The values of given function are shown in the below table.

Step-by-step explanation:

The given function is

$f(x)=\frac{x^2-5x}{x^2-x-20}$

Simplify the given function.

$f(x)=\frac{x(x-5)}{x^2-5x+4x-20}$

$f(x)=\frac{x(x-5)}{x(x-5)+4(x-5)}$

$f(x)=\frac{x(x-5)}{(x+4)(x-5)}$

Cancel out the common factor.

$f(x)=\frac{x}{x+4}$

Substitute x=5.5 in the above equation.

$f(5.5)=\frac{5.5}{5.5+4}$

$f(5.5)=\frac{5.5}{9.5}$

$f(5.5)=0.57894736842$

$f(5.5)\approx 0.578947$

Similarly find the value for all values of x.

The values of given function are shown in the below table.

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Step-by-step explanation:

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2 years ago

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BaLLatris [955]

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2 years ago

1. The probability of telesales representative making a sale on a customer call is 0.15.

Mumz [18]

Answer:

$n = 33$

$n = 19$

Step-by-step explanation:

From the question we are told that

The probability of telesales representative making a sale on a customer call is $p = 0.15$

The mean is $\mu = 5$

Generally the distribution of sales call made by a telesales representative follows a binomial distribution

i.e

$X \~ \ \ \ B(n , p)$

and the probability distribution function for binomial distribution is

$P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}$

Here C stands for combination hence we are going to be making use of the combination function in our calculators

Generally the mean is mathematically represented as

$\mu = n* p$

=> $5= n * 0.15$

=> $n = 33$

Generally the least number of calls that need to be made by a representative for the probability of at least 1 sale to exceed 0.95 is mathematically represented as

$P( X \ge 1) = 1 - P( X < 1 ) > 0.95$