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the average growth hair is 0.5 inch per month. find how long it take a human to grow 3 inches of hair. write an eqution and solve

Answers

Answer 1

Answer:

Answer:

x+ 0.5=3 this is the equation

Step-by-step explanation:

first you would subtract 0.5 on both sides. Then You have an answer of x=2.5

Hope this helped have a good day :D

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If the triangles are congruent,then x = [?]134x+12x+y8x-2yEnter the number that belongs inthe green boxEnter
The blood platelet counts of a group of women have a bell-shaped distribution with a mean of 261.5 and a standard deviation of 67.4 . (All units are 1000 cells/μL.) Using the empirical rule, find each approximate percentage below. a. What is the approximate percentage of women with platelet counts within 1 standard deviation of the mean, or between 194.1 and 328.9 ? b. What is the approximate percentage of women with platelet counts between 59.3 and 463.7 ?

How many grains of sand are there in 6300kg of sand? Give your answer in standard from

Answers

Answer:

5 g

Step-by-step explanation:

Answer:

16,191,000,000

Step-by-step explanation:

1 kg= 2.57 millions grains of sand

2.57 million*6300= 16,191,000,000

If the level of significance of a hypothesis test is raised from .01 to .05, the probability of a Type II error a. will not change. b. will increase. c. will also increase from .01 to .05. d. will decrease.

Answers

Answer:

d. Decrease

Step-by-step explanation:

A Type II error is when we fail to reject a false null hypothesis. Higher values of α make it easier to reject the null hypothesis, so choosing higher values for α can reduce the probability of a Type II error.

The consequence here is that if the null hypothesis is true, increasing α makes it more likely that we commit a Type I error (rejecting a true null hypothesis).

So using lower values of α can increase the probability of a Type II error.

Final answer:

Raising the level of significance in a hypothesis test from .01 to .05 would decrease the probability of making a Type II error. This is because as we become more accepting of risk in making a Type I error, we simultaneously reduce the risk of making a Type II error.

Explanation:

The level of significance in a hypothesis test is the probability that we are willing to accept for incorrectly rejecting the null hypothesis or making a Type I error. If the level of significance is raised, there is a higher chance we incorrectly reject the null hypothesis, increasing the chances of a Type I error. However, this also has an effect on the probability of committing a Type II error, which is to incorrectly accept the null hypothesis.

Specifically, when the level of significance of a hypothesis test is raised from .01 to .05, the probability of a Type II error (option b) will decrease. The reason for this is that increasing the level of significance or alpha means we are more likely to reject the null hypothesis. As we are more accepting of risk in terms of making a Type I error, we are less likely to make a Type II error, as the two error types often move in opposite directions. Thus, the answer to your question is d. The probability of a Type II error will decrease if the significance level is raised from .01 to .05.

Learn more about Hypothesis Testing here:

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Nondigestible carbohydrates can be used in diet foods, but they may have Problem 7.2 effects on colonic hydrogen production in humans. We want to test to see if inulin, fructooligosaccharide, and lactulose are equivalent in their hydrogen production. Preliminary data suggest that the treatment means could be about 45, 32, and 60 respectively, with the error variance conservatively estimated at 35. How many subjects do we need to have power .95 for this situation when testing at the ????1 = .01 level?

Answers

Answer:

null hypothesis = µ1=µ2=µ3; µ1= popuation mean of inulin µ2 = population mean of fructicoligosaccharide =µ3=population mean of lactulose

alternative hypothesis =µ1≠µ2 ≠µ3

t1= µ-45/s1 / N-1

at the significance level 0.01

t at ᵅ/2 =0.005 and degree of freedom= 35-1=34 is 2.25

t1 = µ-32/s1/N-1

2.25= µ-45/s1/34

= s1/34= µ-45/2.25

=s1 =(µ-45/2.25)*34

t2= µ-32/s2/34

2.25 =µ-32/s2/34

s2/34= µ-32/2.25

s2=µ-32/2.25*34

T= 45-32/s1/sqrt34+s2/sqrt34

t at 0.005 and no of grees of freedom 68 =2.37

2.37=45-32/

or s1/sqrt34+s2/sqrt34 = 13/2.37

s1+s2 = 13/2.37 *5.83

s1+s2= 31.98 or 32

(µ1-45/2.25)*34+µ2-32/2.25*34=32

or µ1+µ2 = 2525

µ1=2525-µ2

µ1 and µ2 are not equal

thus null hypothesis is rejected

conclusion all the three components are not in equal amount in hydrogen production

Explanation : in this experiment we have to prove whether the means of insulin,fructicoligosaccharide and lactulose are equal. so even if we prove that two of them are not equal null hypothesis will be rejected. we use student-t test because we have to compare the means of two population.

NEED THIS QUESTION ASAP!

Answers

Answer:

1. The last one

2. The third one

How many randomly selected employers must we contact in order to create an estimate in which we are 95% confident with a margin of error of 9%? b) Suppose we want to reduce the margin of error to 4%. What sample size will suffice? c) Why might it not be worth the effort to try to get an interval with a margin of error of 1%?

Answers

Answer:

a)n=543

b)n=1509

c)n=13573

Step-by-step explanation:

c=98%,

$E=0.05$

Margin Error $E=Zα/2√p(1-p)/n$

but $n=((Zα/2)/n)²×p(1-p)$

where the confidence level is 1-α=0.98

cross multiply

$Zα/2=2.33$

where p=0.5

input the values

$n=(2.33/0.55)²×0.5(1-0.5)=543$

n=0.33

b) E=0.33

$E=Zα/2√p(1-p)/n$

$n=((Zα/2)/n)²×p(1-p)$

$1-α=0.01$ confidence level

n=(2.33/0.33)²×0.5(1-0.5)=1504

$n=1504$

c) $E=Zα/2√p(1-p)/n$

$n=((Zα/2)/n)²×p(1-p)$

1-α=0.98

cross multiply

Zα/2=2.33

p=0.5

n=(2.33/0.01)²×0.5(1-0.5)=13573

$n=13573$

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Answers

9514 1404 393

Answer:

maximum height: 26.5 ft
air time: 2.5 seconds

Step-by-step explanation:

I find the easiest way to answer these questions is to use a graphing calculator. It can show you the extreme values and the intercepts. The graph below shows the maximum height is 26.5 ft. The time in air is about 2.5 seconds.

If you prefer to solve this algebraically, you can use the equation of the axis of symmetry to find the time of the maximum height:

t = -b/(2a) = -(40)/(2×-16) = 5/4

Then the maximum height is ...

h(5/4) = -16(5/4)² +40(5/4) +1.5 = -25 +50 +1.5 = 26.5 . . . feet

Now that we know the vertex of the function, we can write it in vertex form:

h(t) = -16(t -5/4)² +26.5

Solving for the value of t that makes this zero, we get ...

0 = -16(t -5/4)² +26.5

16(t -5/4)² = 26.5

(t -5/4)² = 26.5/16 = 1.65625

Then ...

t = 1.25 +√1.65625 ≈ 2.536954