If a triangle has sides 12, 13 and 16, is it a right triangle? Show your work

Answers

Answer 1

Answer:

yes it is right triangle

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What is the GCF of 4x^2 - 100? I'm not sure if there is even a GCF.

Answers

The greatest common factor is 2.

In a recent Super Bowl, a TV network predicted that 31 % of the audience would express an interest in seeing one of its forthcoming television shows. The network ran commercials for these shows during the Super Bowl. The day after the Super Bowl, and Advertising Group sampled 120 people who saw the commercials and found that 40 of them said they would watch one of the television shows. Suppose you have the following null and alternative hypotheses for a test you are running:1. H0: p = 0.53 Ha: p ≠ 0.53
2. H0: p = 0.53 Ha: p ≠ 0.53
Calculate the test statistic, rounded to 3 decimal places.

Answers

Answer:

-4.317

Step-by-step explanation:

The z test statistic for testing of 1-proportion can be computed as

$z=\frac{phat-p}{\sqrt{(pq)/(n) } }$

We know that

phat=x/n.

We know that x=40 and n=120.

Thus,

phat=40/120=0.3333

p=hypothesized proportion=0.53

q=1-p=1-0.53=0.47

So, required z-statistic is

$z=\frac{0.3333-0.53}{\sqrt{(0.53(0.47))/(120) } }$

$z=(-0.1967)/( 0.04556 )$

z=-4.317.

Thus, the required test statistic value for given hypothesis is z=-4.317.

Solve for:-9a = 27
please someone help!!!

Answers

Answer:

a = -3

:)))))))))))))))

Calcular: log 4 2 = (escribir el resultado como decimal)

Answers

Answer:

1/2 or 0.5

Step-by-step explanation:

Recent research suggests that depression significantly increases the risk of developing dementia later in life (BBC News, July 6, 2010). In a study involving 949 elderly persons, it was reported that 22% of those who had depression went on to develop dementia, compared to only 17% of those who did not have depression.a. Choose the relevant population and the sample. (You may select more than one answer. Bold the answer(s) that are correct.)i. The sample consists of all elderly people.ii. The sample consists of 949 elderly people.iii. The population is all younger and elderly people.iv. The population is all elderly people.b. Do the numbers 22% and 17% represent the population parameters or the sample statistics? Bold the answer that is correct.i. Population parametersii.Sample Statistics

Answers

Answer:

a. ii. and iv.

b. ii.

Step-by-step explanation:

Hello!

The research study involves elderly persons it was divided into two considering those who suffered depression and those that didn't suffer depression, then the proportion of persons that suffered dementia was counted.

a. Choose the relevant population and the sample.

i. The sample consists of all elderly people.

ii. The sample consists of 949 elderly people.

iii. The population is all younger and elderly people.

iv. The population is all elderly people.

Note: When you determine a population from a text, you NEVER mention a number, even if the population is finite and the total of individuals in it is known. When you determine a sample it always comes with its size.

b. Do the numbers 22% and 17% represent the population parameters or the sample statistics?

i. Population parameters

ii.Sample Statistics

The proportions were determined after taking the sample, as said in the text, first the sample was taken, then the researchers determined from this sample wich elders suffered from depression and which didn't suffer from depression and these two groups, the number of people with dementia was determined. Meaning these are sample values, not population parameters.

I hope it helps!

Suppose that 2121% of all births in a certain region take place by Caesarian section each year. a. In a random sample of 500500 births, how many, on average, will take place by Caesarian section? b. What is the standard deviation of the number of Caesarian section births in a sample of 500500 births? c. Use your answers to parts a and b to form an interval that is likely to contain the number of Caesarian section births in a sample of 500500 births.

Answers

Answer:

a) $E(X) = np=500*0.21= 105$

b) $Sd(X) = √(82.95)= 9.108$

c) Assuming a the normality assumption we will have within 2 deviations from the mean most of the data from the distribution and the interval for this case would be:

$\mu -2\sigma = 105-2*9.108=86.785$

$\mu +2\sigma = 105+2*9.108=123.215$

So we expect about 86 and 123 most of the numbers of Caesarian section births

Step-by-step explanation:

For this case we can define the random variable X as the number of births in the Caesarian section and from the data given we know that the distribution of X is:

$X \sim Binom (n = 500, p=0.21$

Part a

The expected value for this distribution is given by:

$E(X) = np=500*0.21= 105$

Part b

The variance is given by:

$Var(X) = np(1-p) = 500*0.21*(1-0.21)= 82.95$

And the deviation would be:

$Sd(X) = √(82.95)= 9.108$

Part c

Assuming a the normality assumption we will have within 2 deviations from the mean most of the data from the distribution and the interval for this case would be:

$\mu -2\sigma = 105-2*9.108=86.785$

$\mu +2\sigma = 105+2*9.108=123.215$

So we expect about 86 and 123 most of the numbers of Caesarian section births

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Ten measurements of impact energy on specimens of A238 steel at 60 ºC are as follows: 64.1, 64.7, 64.5, 64.6, 64.5, 64.3, 64.6, 64.8, 64.2, and 64.3 J. a. Use the Student’s t distribution to find a 95% confidence interval for the impact energy of A238 steel at 60 ºC. b. Use the Student’s t distribution to find a 98% confidence interval for the impact energy of A238 steel at 60 ºC.