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vazorg [7]
3 years ago
14

Apply the distributive property to create an equivalent expression.

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
8 0
Add the liked terms & un lined terms &.7.:!/‘skass then solve
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Factorize (4d-3e)2 +6d(4d-3e)​
Natasha_Volkova [10]

Answer:

(4d - 3e)(10d - 3e)

Step-by-step explanation:

(4d - 3e)² + 6d(4d - 3e) ← factor out (4d - 3e) from each term

= (4d - 3e)(4d - 3e + 6d) ← collect like terms inside parenthesis

= (4d - 3e)(10d - 3e)

4 0
2 years ago
Read 2 more answers
[PICTURE ATTACHED] sigh, another variability one. if u get it correct, i’ll make you brainliest! :’)
Dominik [7]

Answer: B

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Pls help me I promise I’ll mark u brainleiest if I type in the fastest answer
Sever21 [200]

Answer:

1. $-50

2. $25

Step-by-step explanation:

1. If she had $150 in her bank account and bought a bike for $200, then that means she spent all of her money PLUS $50 extra then what she had.  That means $200-$150=$50.  Her $150 is spent and that $50 becomes negative because she paid $200 when she only had $150.

2.  If she deposits $75 in her account then it will be $75+(-50).  That translates to $75-$50 which is $25.  

+ and - = -

+ and + = +

- and - = +

4 0
3 years ago
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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
Based on the table of values below find the slope between points where x=2 and where x=6
kobusy [5.1K]
Slope=(12-4)/(2-6)=-2
7 0
3 years ago
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