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valkas [14]
3 years ago
15

(y+6)^2-(y-2)^2 I got 16y+26 but it is wrong?

Mathematics
1 answer:
Gemiola [76]3 years ago
5 0

Answer:

16y + 32

Step-by-step explanation:

Expand each term.

(y+6)² - (y-2)²

= (y+6)(y+6) - (y-2)(y-2)

= y² + 12y + 36 - (y² - 4y + 4)

Subtract the second group by changing each term's signs

= y² + 12y + 36 - y² + 4y - 4

Collect like terms

= 16y + 32

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Write in y= mx + b form: a line through (-3,1) having the same y-intercept as the graph of x-2y=-4
MariettaO [177]

Step-by-step explanation:

Aight, so the same intercept

- 2y =  - 4 - x =  =  =  >  \\ y =  \frac{1}{2} x + 2

m=½

y =  \frac{1}{2} x + b =  =  =  >  \\ now \: let \: us \: replace \: the \: point \\ 1 =  \frac{1}{2} ( - 3) + b =  =  =  >  \\  \frac{5}{2}  = b

soooo

y =  \frac{1}{2} x +  \frac{5}{2}

7 0
1 year ago
2) AT&T charges an initial fee of $5.80 plus $0.30 per text. If Izzy has her cell phone
balu736 [363]

Answer: 50 texts

Step-by-step explanation: 20.8 - 5.80 = 15

15 ÷ .30 = 50

3 0
3 years ago
Draw the graph of R = {(1,1), (2,2), (3,3), (4,4), (5,5), (1,4), (2,5), (4,1), (5,2)}
tresset_1 [31]

Answer:

Step-by-step explanation:

3 0
1 year ago
Which list of ordered pairs represents solutions to x + y = 2 ?
Sloan [31]
(-4, 6), (0, 2), (4, -2) are all ordered pairs that have the sum as 2.
-4+6 is the same thing as 6-4 if there's more positive than negative 
0+2=2 because 0 plus a number equals to that number
4-2=2 because you're adding a negative number with the positive number together to get the answer

Just be aware that I might not make any sense at all

3 0
3 years ago
For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df
rjkz [21]

Answer:

a) For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

b) For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

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

Part a

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

Part b

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

3 0
3 years ago
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