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3 years ago

6

Locate the prepositional phrase and type it in the blank. Then, identify the function of the phrase. Above the door hung a horse

shoe. prepositional phrase: function of phrase: (Type adjective or adverb ).

1 answer:

olga_2 [115]3 years ago

3 0

“Above the door” is the prepositional phrase but I can’t help you you it’s the second part, sorry.

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May I get help on these questions, please?

Virty [35]

Hello! here are the basic rules that you need to know to solve these problems:

a diameter is 2 × the radius
the circumference rule is 2×π×radius

I will solve the first question to show how to format.

1: 5/2=2.5
2×2.5×π=5×π=15.7

5 0

2 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

calvin and 4 of his friends want to share 4 pounds of nuts equally.write an expression to show the fraction of the nuts each fri

PilotLPTM [1.2K]

4/5

The 4 is how many pounds of nuts there are and dividing by 5 shows that you are sharing the nuts equally amongst the 5 friends.

5 0

3 years ago

Given= 3a + 5 = 17

choli [55]

Answer:

a=4

Step-by-step explanation:

you subtract the 5 from both sides and then you divide the 3 into the three and then you have 4

5 0

4 years ago

Read 2 more answers

!! AP CALCULUS AB !!<br> find the limit of [(secx*sinx) + (cscx*cosx)]/(3secx) as x approaches 3pi/2

uysha [10]

$f(x)= \frac{\frac{\sin{x}}{\cos{x}}+\frac{\cos{x}}{\sin{x}}}{3\sec{x}}$

$\frac{\frac{\sin{x}}{\cos{x}}+\frac{\cos{x}}{\sin{x}}}{3\sec{x}} \implies \\ \frac{\sin^2{x}+\cos^2{x}}{\sin{x}\cos{x}}\frac{\cos{x}}{3}\implies \\\frac{1}{\sin{x}\cos{x}}*\frac{\cos{x}}{3}\implies \\ \frac{1}{3\sin{x}}$

So,

$\lim_{x \rightarrow \frac{3\pi}{2}}{f(x)=f(\frac{3\pi}{2})=\frac{1}{3\sin{\frac{3\pi}{2}}}=-\frac{1}{3}$

6 0

3 years ago