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3 years ago

Solve for x x² + 10x + 25

Mathematics

2 answers:

Sonbull [250]3 years ago

7 0

Hi,

Answer: (x+5)^2

My work: For this problem can be easily achieved by factoring your terms. To do this you figure out what can go into 10x and 25 which is 5. From the there you take x^2 and 10x and see what can take out which would be x. Your answer would be (x + 5)^2 or (x +5) (x + 5). This can be done in 2 easy steps!

Numerical work:

1.x^2 10x +25

Before this step figure out what goes into you equation.

2. (X + 5)^2 or (X+5) (X + 5)

Galina-37 [17]3 years ago

5 0

Let's factor x2+10x+25

x2+10x+25

The middle number is 10 and the last number is 25.

Factoring means we want something like

(x+_)(x+_)

Which numbers go in the blanks?

We need two numbers that...

Add together to get 10

Multiply together to get 25

Can you think of the two numbers?

Try 5 and 5:

5+5 = 10

5*5 = 25

Fill in the blanks in

(x+_)(x+_)

with 5 and 5 to get...

(x+5)(x+5)

Answer:

(x+5)(x+5)

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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 $t_{\alpha}= -2.571$

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3 years ago

What are the solutions to the equation 9x2 = 49, rounded to the nearest tenth?

Sergio [31]

I would go for A if anything

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3 years ago

Read 2 more answers

witch rule can you use to find the nth term of an arithmetic sequence in witch the common difference is 5 and a12=63

rewona [7]

Any arithmetic sequence can be expressed as:

a(n)=a+d(n-1), a=initial term, d=common difference, n=term number

We are only told that d=5 and a(12)=63 so:

53=a+5(12-1)

53=a+60-5

53=a+55

a=-2 now we have solved for a and we can say:

a(n)=-2+5(n-1) which neatened up....

a(n)=-2+5n-5

a(n)=5n-7

The above is the rule to find the value of the nth term.

3 0

3 years ago