Solve the following problem: X^4-7x^2-30=0

Juanita says that 3.5[4d - (2) (1.5)] and 2[7d - (5)(1.05)] are equivalent. Is Juanita correct? Explain your reasoning.

valkas [14]

Answer:

Yes

Step-by-step explanation:

3.5 [4d - 2*1.5] = 3.5*4d - 3.5*2*1.5

= 14d - 10.5

2*[7d - 5*1.05] = 2*7d - 2*5*1.05

= 14d - 10.5

When we are simplifying both the expression using distributive property, we get the same expression at the end.

8 0

3 years ago

Write an algebraic expression for the following: 17 more than a number.

Marysya12 [62]

Answer:

x+17

Step-by-step explanation:

Writing this for the sake of adding more characters, but yeah. 'A number' is basically x, add 17....bam! x+17

6 0

2 years ago

Read 2 more answers

Cate and Elena were playing a card game. The stack of cards in the middle had 24 cards in it to begin with. Cate added 8 cards t

Yuki888 [10]

You start with 24, there are 8 added, there are 12 taken away, then 9 are taken away. The equation looks like:
24 + 8 - 12 - 9 = <u>11</u>

7 0

3 years ago

Input output please

dlinn [17]

Step-by-step explanation:

89
81
50
83
295

<h2>hope it helps.</h2><h2>stay safe healthy and happy.</h2>

8 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