Find the length of side x.

A triangle has a side lengths of 3 and 8. The diagonal has a length of 11. Can this be a right triangle? Show your work to prove

Grace [21]

If the triangle is a right triangle, the pythagorean theorem would work
since it doesn’t, the triangle is NOT A RIGHT TRIANGLE

(see attached pic for work :))

6 0

3 years ago

Someone help please I wi appreciate itttt so much

Dmitry_Shevchenko [17]

Answer:

65

Step-by-step explanation:

66

3 0

3 years ago

An adult blue whale can eat about 40 million Krill a day write this number in a scientific notation

BigorU [14]

4x10^7 The number of zeros the given number has would be the exponent

7 0

3 years ago

Read 2 more answers

Help ASAP please please 14 and 15

Yuliya22 [10]

I think 14 is B (idrk) and 15 is definitely C

8 0

3 years ago

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .

Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b) "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e) "=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f) "=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c. Central area =.99, df = 20

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d. Central area =.99, df = 50

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means $\alpha =0.01$

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means $\alpha =0.025$

We can use the following excel code:

"=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

8 0

3 years ago