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4 years ago

The function g is defined by g(x) = 2x - 4. Find g(z-3).

Mathematics

1 answer:

borishaifa [10]4 years ago

6 0

Answer:

2z -10

Step-by-step explanation:

g(x) = 2x - 4.

Let x =z-3

g(z-3) = 2(z-3) - 4.

Distribute

g(x-3) = 2z -6 -4

=2z -10

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What is the measure of ZXZW?

ahrayia [7]

Answer:

Step-by-step explanation:

if you look inside the triangle,

you have one angle=65

the acute angle=180-150=30

<YZX=180-(65+30)=85

the angle <xzw is its supplement, 180-85=95

8 0

3 years ago

Add:<br> 1⁄2 + 2⁄5 + 1⁄4 =

Finger [1]

$\frac{23}{20}$

5 0

3 years ago

Read 2 more answers

These are the first four terms of a sequence. 3 -1 -5 -9. Find the nth term.

podryga [215]

Answer:

4n-1

Step-by-step explanation:

Diffrence bweetween each number, 3 and -1, -1 and -5, -5 and -9 is 4

find the closest numbers

4*1=4

4*0=0

4*-1=-4

4*-2=-8

now we figure out how to make these numbers into our terms of this sequence

4n-1

4*1-1=3

4*0-1=-1

4*-1-1=-3

4*-2-1=-9

so the nth term is 4n-1

4 0

2 years ago

The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

vfiekz [6]

Answer:

a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. 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.

In this problem we have that:

$p = 0.75$

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X = 9)$ when $n = 12$ . So

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

$P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581$

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X \geq 16)$ when $n = 20$

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

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

$P(X = 16) = C_{20,16}.(0.75)^{16}.(0.25)^{4} = 0.1897$

$P(X = 17) = C_{20,17}.(0.75)^{17}.(0.25)^{3} = 0.1339$

$P(X = 18) = C_{20,18}.(0.75)^{18}.(0.25)^{2} = 0.0669$

$P(X = 19) = C_{20,19}.(0.75)^{19}.(0.25)^{1} = 0.0211$

$P(X = 20) = C_{20,20}.(0.75)^{20}.(0.25)^{0} = 0.0032$

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1897 + 0.1339 + 0.0669 + 0.0211 + 0.0032 = 0.4148$

c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.

The expected value of the binomial distribution is:

$E(X) = np$

In this question, we have $n = 22$

$E(X) = 22*0.75 = 16.5$

The closest integer to 16.5 is 17.

7 0

3 years ago

1. A company has a cash portfolio measured in millions. The drift is 0.1 per month, variance is 0.16per month. The initial cash

Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

From the information given:

The probability distribution at the end of 6 months is determined as follows:

After 6 months;

Mean of probability distribution = value of Initial cash + $\alpha$ T

=2.0 +(0.1 × 6)

=2.6

After 6 months;

The probability distribution's standard deviation is estimated by using the following formula:

Standard deviation:

$= b\sqrt{T}$

$= 0.4 \times \sqrt{6}$

= 0.9798

Hence, after 6 months;

The company's cash position is supposed to be allocated monthly, with the following expenses.

Mean 2.6

Standard deviation 0.9798

Variance 0.96

After 12 months, the probability distribution is as follows:

Mean = value of Initial cash + $\alpha$ T

= 2.0 +(0.1 × 12)

= 3.2

The standard deviation is:

The standard deviation of probability distribution = $b \sqrt{T}$

$= 0.4 \times \sqrt{12}$

= 1.3856

Hence, after 6 months;

The company's cash position is supposed to be allocated monthly, with the following expenses.

Mean 3.2

Sandard deviation 1.3856

Variance 1.92

in 6-month distribution, the probability of the negative value of the cash position is as follows.

Now, for us to find the negative cash distribution;

We need to estimate the z -scores value.

The z-score inform us greatly on the concept of how far a particular data point is from the mean.

For a normal distribution;

$z = \dfrac{x-\mu}{\sigma}$

Here;

the value of x = zero as a result that if it exceeds zero. the cash position will be negative.

∴

$z = \dfrac{x-\mu}{\sigma}$

$z = \dfrac{0 - 2.6}{0.9798}$

$z = -2.6536$

Using the standard distribution tables, it is now possible to calculate that the likelihood N(-2.65) equals 0.004 or 0.4 percent.

As a result, there's a 0.4 percent chance of getting a negative cash balance after six months.

For 12 months distribution:

The Probability of negative cash position is calculated as follows:

$z = \dfrac{x-\mu}{\sigma} \\ \\ z = \dfrac{0-3.2}{1.3856} \\ \\ z = -2.3094$

Using the standard distribution tables,

N(-2.31) equals 0.0104 or 1.04 percent.

As a result, there's a 1.04 percent chance of getting a negative cash balance after 1 year

c) To determine the time period over which the likelihood of achieving a negative cash condition is highest, it's necessary to examine the z-score more closely. Essentially, the z-score measures the difference between a given value(x) and the mean of all potential values $(\mu)$ , expressed in terms of the total set's standard deviation $(\sigma)$

This suggests that the higher the z-score, the greater the difference occurring between x and $\mu$ , and thus the likelihood of receiving x is minimal. As a result, the best chance of finding a certain value is when the z-score is the lowest.

To do so, calculate the derivative of the z-score in relation to the time interval. The point where the derivative is equivalent to zero is where the z-scores are at their lowest.

The first step is to go over the z-score formula in more detail, as seen below.;

$z = \dfrac{x-\mu}{\sigma} \\ \\ z = \dfrac{0-(initial \ value + \alpha T)}{b \sqrt{T}} \\ \\ z = \dfrac{-initial \ value }{b\sqrt{T}}-\dfrac{a \sqrt{T}}{b} \\ \\$

Now, compute the derivative of this equation with respect to T as follows:

$\dfrac{dz}{dT}= \dfrac{initial \ value \times T^{-\dfrac{3}{2}}}{2b} - \dfrac{aT^{-\dfrac{1}{2}}}{2b}$

Now, figure out the value of T at which this derivative is equal to zero by substituting all values as follows:

$0 = \dfrac{2.0 \times T^{-\dfrac{3}{2}}}{2\times 0.4}- \dfrac{0.1 \times T^{-\dfrac{1}{2}}}{2 \times 0.4} \\ \\ \\ 0.1 \times T^{-\dfrac{1}{2}}= 2.0 \times T^{-\dfrac{3}{2}} \\ \\ \\T = \dfrac{2}{0.1} \\ \\ \\ T = 20$

As a result, the time period in which achieving a negative cash condition is = 20 months.

3 0

3 years ago