Help please..........

Three functions are given below: f(x), g(x), and h(x). Find the axis of symmetry for each function, and rank the functions based

m_a_m_a [10]

Answer:

Step-by-step explanation:

f(x)=-4(x-2)²+3

put x-2=0

axis of symmetry is x=2

g(x)=-2(x²+8x)+15

=-2(x²+8x+4²-4²)+15

=-2(x+4)²+32+15

=-2(x+4)²+47

put x+4=0

axis of symmetry is x=-4

h(x) is minimum at x=1

axis of symmetry is x=1

3 0

3 years ago

The CEO of a large corporation asks his Human Resource (HR) director to study absenteeism among its executive-level managers at

Fynjy0 [20]

Answer:

Following are the answer to this question:

Step-by-step explanation:

Given:

n = 30 is the sample size.

The mean $\bar X$ = 7.3 days.

The standard deviation = 6.2 days.

df = n-1

$= 30-1 \\ =29$

The importance level is $\alpha$ = 0.10

The table value is calculated with a function excel 2010:

$= tinv (\ probility, \ freedom \ level) \\= tinv (0.10,29) \\ =1.699127\\ = t_{al(2x-1)}= 1.699127$

The method for calculating the trust interval of 90 percent for the true population means is:

Formula:

$\bar X - t_{al 2,x-1} \frac{S}{\sqrt{n}} \leq \mu \leq \bar X+ t_{al 2,x-1} \frac{S}{\sqrt{n}}$

$=\bar X - t_{0.5, 29} \frac{6.2}{\sqrt{30}} \leq \mu \leq \bar X+ t_{0.5, 29} \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 \frac{6.2}{\sqrt{30}}\leq \mu \leq7.3 +1.699127 \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 (1.13196)\leq \mu \leq7.3 +1.699127 (1.13196) \\\\=5.37 \leq \mu \leq 9.22 \\$

It can rest assured that the true people needs that middle managers are unavailable from 5,37 to 9,23 during the years.

7 0

3 years ago

A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true prop

attashe74 [19]

Answer:

$0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286$

$0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562$

The 99% confidence interval would be given by (0.286;0.562)

Step-by-step explanation:

Information given:

$X= 36$ represent the families owned at least one DVD player

$n= 85$ represent the total number of families

$\hat p=\frac{36}{85}= 0.424$ represent the estimated proportion of families owned at least one DVD player

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286$

$0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562$

The 99% confidence interval would be given by (0.286;0.562)

8 0

3 years ago

For each of the following binomial random variables, specify n and p. (a) A fair die is rolled 50 times. X = number of times a 5

Keith_Richards [23]

Answer:

a) $n = 50, p = \frac{1}{6}$

b) $n = 16, p = \frac{1}{100}$

c) $n = 26, p = 0.25, \mu = 6.5$

Step-by-step explanation:

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 combinatios 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.

(a) A fair die is rolled 50 times. X = number of times a 5 is rolled

The die is rolled 50 times, so $n = 50$ .

Each roll can have 6 outcomes. So the probability that 5 is rolled is $p = \frac{1}{6}$

(b) A company puts a game card in each box of cereal and 1/100 of them are winners. You buy sixteen boxes of cereal, and X = number of times you win.

You buy 16 boxes of cereal, so $n = 16$ .