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gogolik [260]
3 years ago
13

HELP ASAP PLEASE & SHOW UR WORK!! THANK YOU <3

Mathematics
1 answer:
olchik [2.2K]3 years ago
4 0

Answer:

x^2 - 5x = 14

x^2 - 5x-14 = 0

Now, we have to find the multiples of -14 and numbers that add/subtract to -5; this helps to find the factors.

-7+2 = -5

-7*2 = -14

Based on the above info, we can find out my factors; they are:-

(x-7)(x+2)

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traingle ABC has been dilated aboit point A by a scale factor of 1/3. What are the lengths, in units, of the three sides of tria
solniwko [45]

Answer:

You need to show the graph so i can see numbers, you cant dilate the traingle without knowing where it is on the graph.

Step-by-step explanation:

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Answer: Jason can imagine everyone naked while on the stage, it will make him smile, and feel free.

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3 0
2 years ago
Growth or decay, and determine the percentage rate of increase or decrease.
NNADVOKAT [17]

The expression y = 660(0.902)ˣ represents a decay , the rate of decrease is 9.8% .

In the question ,

it is given that ,

the the exponential function is y = 660(0.902)ˣ

we need to determine that , the equation represents a growth or decay .

0.902 is the variable that will determine a growth or decay .

Since, 0.902 is less than 1 , so it represents a decay function .


To find the rate we subtract 0.902 from 1

= 1 - 0.902

= 0.098

= 9.8% decrease rate

Therefore , The expression y = 660(0.902)ˣ represents a decay , the rate of decrease is 9.8% .

The given question is incomplete , the complete question is

Identify if it's Growth or Decay , and determine the percentage rate of increase or decrease , the function is y = 660(0.902)ˣ ?

Learn more about Growth and Decay here

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4 0
1 year 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
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