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3 years ago

7

A gym has yoga classes. Each class has students. If there are classes, write an equation to represent the total number of studen

ts taking yoga.

1 answer:

zzz [600]3 years ago

7 0

Answer:

s = 14c

Step-by-step explanation:

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!!!!!!!!!!!!help me

steposvetlana [31]

Answer:

Step-by-step explanation:

i'll do a few of them

1) one inch plus a half inch plus an eighth inch plus a sixteenth inch

1 + 1/2 + 1/8 + 1/16 read the tick marks

1 + 8/16 + 2/16 + 1/16 find a common denominators

1 and 11/16 add the sixteenth numerator

2)

3)

4) 6 inch plus an eighth inch plus a sixteenth inch

6 + 1/8 + 1/16 read the tick marks

6 + 2/16 + 1/16 find a common denominators

6 and 3/16 add the sixteenth numerator

5) nine inch plus a quarter inch plus an eighth inch

9 + 1/4 + 1/8 read the tick marks

9 + 2/8 + 1/8 find a common denominators

9 and 3/8 add the sixteenth numerator

5 0

3 years ago

Please help asap

Nostrana [21]

I Think The answer is d I hope it helps My friend Message Me if I’m wrong and I’ll change My answer and fix it for you

6 0

2 years ago

A1 = 7 and r = 5<br> I need this please

Sloan [31]

A=7 is what I got since this was a equation problem

3 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

Which of the following graphs shows the solution for the inequality y - 4 > 2(x + 2)?

Murljashka [212]

Answer:

D. Graph C

Step-by-step explanation:

y - 4 > 2(x + 2)

y > 2x + 4 + 4

y > 2x + 8

Let y = 2x + 8, Then

X-intercept (0, 8)

Y-intercept (-4, 0)

Since the inequality sign is > we use broken line.

to test the direction, take (0, 0)

y > 2x + 8

0 > 0+ 8

0 > 8 Which is false

so, the answer will be above the broken line

6 0

2 years ago

Read 2 more answers