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Irina18 [472]
4 years ago
14

CAN SOMEONE HELP?????????????????????/

Mathematics
1 answer:
Agata [3.3K]4 years ago
4 0

Answer:

see explanation

Step-by-step explanation:

h(- 4) means what is the value of h(x) when x = - 4

Now x = - 4 ≠ 2, thus

h(- 4)

= -  \frac{1}{3} × (- 4)² + 4

= - \frac{1}{3} × 16 + 4

= - \frac{16}{3} + \frac{12}{3} = - \frac{4}{3}

h(2) means what is the value of h(x) when x = 2, thus

h(2) = - 2

h(5) means what is the value of h(x) when x = 5

Now x = 5 ≠ 2, thus

h(5)

= - \frac{1}{3} × 5² + 4

=-  \frac{1}{3} × 25 + 4

= - \frac{25}{3} + \frac{12}{3} =-  \frac{13}{3}

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Find each equation with the correct solution​
stepladder [879]

Answer:

15;9;-8

Step-by-step explanation:

1. 5/3x-3=2/3x+12

x=15 once you combine like terms

2. 5x+35-3x+12=7x+2

2x+47=7x+2

5x=45

x=9

3. 12x+20-3=9x-7

12x+17=9x-7

3x=-24

x=-8

3 0
3 years ago
Can someone solve this? With zeroes please. y = x^2 + 2x - 8
s2008m [1.1K]

Answer:

X=2,-4

Step-by-step explanation:

The roots (zeros) are the  x  values where the graph intersects the x-axis. To find the roots (zeros), replace  y  with  0  and solve for  x .

4 0
3 years ago
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8,326 x 9 =n<br>which is the answer?<br>A. 52,234<br>B.74,934<br>C.2,376<br>D.3008<br>E. none​
Kazeer [188]

Answer:

B

Step-by-step explanation:

3 0
3 years ago
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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

8 0
3 years ago
HELP ASAP YOU GET BRAIBLIEST FOR CORRECT ANSWER
salantis [7]

Answer:

8x+9

if you are happy with my answer, please give brainliest :)

Step-by-step explanation:

Combine like terms:

2x+9+6x=8x+9

7 0
3 years ago
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