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Aleksandr-060686 [28]
3 years ago
7

Select all the rules for the translation of LaTeX: \Delta RSTΔ R S T.

Mathematics
2 answers:
Degger [83]3 years ago
5 0

rasa urrerStep-by-step explanation:

Margaret [11]3 years ago
3 0
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(8p - 2) (6p+2) <br><br> help plz
qaws [65]

Answer:

48p^2+2p-4

Step-by-step explanation:

We can use the FOIL method to expand two multiplied binomials. It states that (a+b)(c+d)=ac+bc+ac+bd. The FOIL method stands for First(first terms) outer(outer terms) inner(inner terms) last(last terms).

So, we can expand our binomials now!

(8p-2)(6p+2)=(8p)(6p)-2(6p)+(8p)(2)-2(2)=48p^2-12p+14p-4=\boxed{48p^2+2p-4}

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Your parents decide that they will help you out the first two months you are at college
vlada-n [284]

Answer:

Step-by-step explanation:

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2 years ago
Pls help asap!!!<br><br><br><br> Solve for x . 2x=y
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The answer to this question is 0
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2 years ago
Determine if the expression – 6y5 z5 – is a polynomial or not. If it is a
sasho [114]

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it is not

Step-by-step explanation:

that is where I truly believe

8 0
2 years ago
Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r
ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean weight = \frac{\sum X}{n} = 21.88 ounces

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 0.201 ounces

            n = sample of boxes = 12

            \mu = population mean weight

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.796 < t_1_1 < 1.796) = 0.90  {As the critical value of t at 11 degrees of

                                                  freedom are -1.796 & 1.796 with P = 5%}  

P(-1.796 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.796) = 0.90

P( -1.796 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.796 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.796 \times {\frac{s}{\sqrt{n} } } , \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ]

                                        = [ 21.88-1.796 \times {\frac{0.201}{\sqrt{12} } } , 21.88+1.796 \times {\frac{0.201}{\sqrt{12} } } ]

                                        = [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0
3 years ago
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