If f(x) = 2x2 +5./(x-2), complete the following statement:
f(0) =

Answers

Answer 1

Answer:

F(6)=82

Step-by-step explanation:

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Just please give me an answer

Factor 6(2p q)^2 -5(2p q) - 25

Answers

treat the 2p+q as x

6x2-5x-25
trial and error
(2x-5)(3x+5)
replace x with (2p+q)
(2(2p+q)-5)(3(2p+q)+5)
(4p+2q-5)(6p+3q+5)

In a study of factors affecting whether soldiers decide to reenlist, 320 subjects were measured for an index of satisfaction. The sample mean is 28.8 and the sample standard deviation is 7.3. Use the given sample data to construct the 98 percent confidence interval for the population mean.

Answers

Answer:

The 98 percent confidence interval for the population mean is between 27.85 and 29.75 subjects.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = (1-0.98)/(2) = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.01 = 0.99$ , so $z = 2.325$

Now, find M as such

$M = z*(\sigma)/(√(n))$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.325(7.3)/(√(320)) = 0.9488$

The lower end of the interval is the mean subtracted by M. So it is 28.8 - 0.9488 = 27.85 subjects.

The upper end of the interval is the mean added to M. So it is 28.8 + 0.9488 = 29.75 subjects.

The 98 percent confidence interval for the population mean is between 27.85 and 29.75 subjects.

You need 2.5 pounds of potatoes. If each potato weighs 5 ounces, how many potatoes will you need?

Answers

Answer:

40 potatoes.

Step-by-step explanation:

16 ounces make a pound, so you multiply that by 2.5 and you get 40.

Final answer:

You need 8 potatoes to total 2.5 pounds, considering each potato weighs 5 ounces.

Explanation:

To solve this problem, you first need to understand that 1 pound is equal to 16 ounces. So, 2.5 pounds would be 40 ounces (16 ounces/pound x 2.5 pounds). If one potato weighs 5 ounces, you would divide 40 (total ounces needed) by 5 (ounces per potato) to find the number of potatoes needed. Thus, you will need 8 potatoes to get approximately 2.5 pounds.

Learn more about Weight Conversion here:

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lled a 12:3:112:3:1 ratio. Such a model can provide the basis for the null hypothesis in a significance test. A cross of white and green summer squash plants gives the number of squash in the second generation F2:131F2:131 white squash, 3434 yellow squash, and 1010 green squash. Are these data consistent with a 12:3:112:3:1 dominant epistatic model of genetic inheritance( white being dominant)? The null hypothesis for the chi‑square goodness‑of‑fit test is

Answers

Here is the full question:

When a species has several variants of a phenotype passed on from generation to generation, we can form a hypothesis about the genetics of the trait based on Mendelian theories of genetic inheritance. For example, in a two-gene dominant epistatic model, the first gene masks the effect of the second gene, leading to the expression of three phenotype variants. Crossing the dominant and recessive homozygote lines would result in a second generation represented by a mix of dominant, intermediate, and recessive phenotype variants in the expected proportions: and respectively, also called a 12:3: 1 ratio.

Such a model can provide the basis for the null hypothesis in a significance test. A cross of white and green summer squash plants gives the number of squash in the second generation F2: 131 white squash, 34 yellow squash, and 10 green squash. Are these data consistent with a 12: 3: 1 dominant epistatic model of genetic inheritance( white being dominant)?

The null hypothesis for the chi-square goodness-of-fit test is

Answer:

The null hypothesis for the chi-square goodness-of-fit test is :

$\mathbf{H_o:p_(white) = (12)/(16), p_(yellow) = (3)/(16); p_(green) = (1)/(16) }$

Step-by-step explanation:

The objective of this question is to state the null hypothesis for the chi-square goodness-of-fit test.

Given that:

There are three colors associated with this model . i,e White , yellow and green and they are in the ratio of 12:3:1

The total number of these color traits associated with this model = 12 + 3 + 1 = 16

Thus ;

The null hypothesis for the chi-square goodness-of-fit test is :

$\mathbf{H_o:p_(white) = (12)/(16), p_(yellow) = (3)/(16); p_(green) = (1)/(16) }$

The daily high temperature in Chicago for the month of August is approximately normal with mean 78 degrees F, and standard deviation 9 degrees F. a. What is the probability that a randomly selected day in August will have a high temperature greater than the mean daily high temperature of 78 degrees F?
b. What is the percentile for a day in August with a high temperature of 75 degrees F?
c. What is the 75th percentile for the daily high temperature for the month of August?
d. What is the interquartile range for the daily high temperature for the month of August?

Answers

Answer:

a) $P(X>78) = P(Z> (78-78)/(9)) = P(Z>0)= 0.5$

b) $P(X<75)= P(Z< (75-78)/(9)) = P(Z<-0.333) = 0.370$

So then 75 F correspond to approximately the 37 percentile

c) $z=0.674<(a-78)/(9)$

And if we solve for a we got

$a=78 +0.674*9=84.07$

So the value of height that separates the bottom 75% of data from the top 25% is 84.07 F.

d) $IQR = 84.07-71.93= 12.14$

See explanation below.

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the daily high temperature in Chicago for the month of August of a population, and for this case we know the distribution for X is given by:

$X \sim N(78,9)$

Where $\mu=78$ and $\sigma=9$

We are interested on this probability

$P(X>78)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=(x-\mu)/(\sigma)$

Using the z score we got:

$P(X>78) = P(Z> (78-78)/(9)) = P(Z>0)= 0.5$

Part b

For this case we can find the percentile with the following probability:

$P(X<75)$

If we use the z score formula we got:

$P(X<75)= P(Z< (75-78)/(9)) = P(Z<-0.333) = 0.370$

So then 75 F correspond to approximately the 37 percentile

Part c

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.25$ (a)

$P(X<a)=0.75$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

$P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.75$

$P(z<(a-\mu)/(\sigma))=0.75$

But we know which value of z satisfy the previous equation so then we can do this:

$z=0.674<(a-78)/(9)$

And if we solve for a we got

$a=78 +0.674*9=84.07$

So the value of height that separates the bottom 75% of data from the top 25% is 84.07 F.

Part d

For this case we know that $IQR = Q_3 - Q_1 = P_(75)-P_(25)$

So then we just need to find the percentile 25.

$P(X>a)=0.25$ (a)

$P(X<a)=0.75$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

$P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.25$