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STatiana [176]
3 years ago
6

I don't know who needs to hear this but I wanted to hopefully reach out to someone.

Mathematics
1 answer:
lora16 [44]3 years ago
8 0

Answer:

oh my- this actually made me happy-

Step-by-step explanation:

You might be interested in
Assume that y varies directly with x then solve y = 2.5 x = .5 when x = 20
velikii [3]

Step-by-step explanation:

y=2.5| ?

x=0.5| 20

It'll be criss-crossed then it'll be:-

2.5×20=50

5 0
2 years ago
The graph shows the value of a certain model of car compared with it's age. which statement is true?
s2008m [1.1K]

The only true statement is A:

"The data show a negative linear relationship."

<h3></h3><h3>Which statement is true?</h3>

On the graph, we can see how the car's vale decreases almost linearly with the age of the car.

Where the response variable would be the one on the y-axis, which is the car's value.

For that linear behavior, we know that there is a correlation coefficient different than zero. So options B, C, and D are false.

Finally, we already saw the linear behavior (decreasing, so the slope is negative). Then we conclude that the only true statement is A.

If you want to learn more about data sets:

brainly.com/question/4219149

#SPJ1

6 0
2 years ago
Which are compostie number 49, 35, 32, 41, 47
tensa zangetsu [6.8K]

Answer: 49, 35, and 32

Step-by-step explanation:

7 * 7 = 49

7 * 5 = 35

4 * 8 = 32

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
Factor the expression completely.<br> 63x – 9
liraira [26]

Answer:

The answer is 9(7x – 1).

Step-by-step explanation:

63x – 9

3 × 3 × 7 × x – 3 × 3

9(7x – 1)

Thus, The answer is 9(7x – 1).

 

<u>-TheUnknownScientist</u><u> 72</u>

4 0
3 years ago
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