NEED HELP NOW FUNCTIONS DONT SKIP

Am I correct? If not please explain why!

Kipish [7]

yes, you are correct. every input corresponds to only one output

6 0

3 years ago

The city of Oakland is experiencing a movement of its population to the suburbs. At present 85% of the total population lives in

Oxana [17]

Answer:

A. $\left[\begin{array}{cc}0.93&0.07&\\0.01&0.99&\end{array}\right]$

B. (0.85 0.15)

C. 79.2% population in the city while 20.8% population in the suburb

Step-by-step explanation:

(a) The transition matrix for the information is

C S

$\left[\begin{array}{cc}0.93&0.07&\\0.01&0.99&\end{array}\right]$

(b) the probability vector for the information is

$(\frac{85}{100} \ \ \frac{15}{100} )$

and this gives us

(0.85 0.15)

(c) we simply multiply the above two matrices to find the percent of the population can be expected to be in each category after one year after

$\left[\begin{array}{cc}0.85&0.15\end{array}\right] \left[\begin{array}{cc}0.93&0.07&\\0.01&0.99&\end{array}\right] = \left[\begin{array}{cc}0.792&0.208&\end{array}\right]$

in the city there are 79.2% while in the suburb, there are 20.8%

5 0

3 years ago

An office that dispenses automotive license plates has divided its customers into categories to level the office workload. Custo

grigory [225]

Answer:

UNIF(2.66,3.33) minutes for all customer types.

Step-by-step explanation:

In the problem above, it was stated that the office arranged its customers into different sections to ensure optimum performance and minimize workload. Furthermore, there was a service time of UNIF(8,10) minutes for everyone. Since there are only three different types of customers, the service time can be estimated as UNIF(8/3,10/3) minutes = UNIF(2.66,3.33) minutes.

6 0

3 years ago

Questions about rational and irrational numbers

FinnZ [79.3K]

Rational numbers are those numbers that are integers and can be expressed in the form of x/y where both numerator and denominator are integers whereas irrational numbers are those numbers which cannot be expressed in a fraction. ... The denominator of a rational number is a natural number(a non-zero number).

Rational numbers and irrational numbers are mutually exclusive: they have no numbers in common. Furthermore, they span the entire set of real numbers; that is, if you add the set of rational numbers to the set of irrational numbers, you get the entire set of real numbers.

Answer: If a number can be written or can be converted to p/q form, where p and q are integers and q is a non-zero number, then it is said to be rational and if it cannot be written in this form, then it is irrational.

4 0

3 years ago

A maker of a certain brand of low-fat cereal barsclaims that the average saturated fat content is 0.5gram. In a random sample of

finlep [7]

Answer:

$t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386$

$p_v =2*P(t_{(7)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance. So then the claim makes sense

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=0.475$

The sample deviation calculated $s=0.183$

We need to conduct a hypothesis in order to check if the true mean is 0.5 or no, the system of hypothesis would be:

Null hypothesis: $\mu = 0.5$

Alternative hypothesis: $\mu \neq 0.5$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=8-1=7$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(7)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance. So then the claim makes sense

4 0

3 years ago

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