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ra1l [238]
3 years ago
5

Solve the equation below. Show all steps for full points. 7r + 13 = 7 + 6r

Mathematics
2 answers:
erma4kov [3.2K]3 years ago
7 0

Answer:

r = - 6

Step-by-step explanation:

Given

7r + 13 = 7 + 6r ( subtract 6r from both sides )

r + 13 = 7 ( subtract 13 from both sides )

r = - 6

ioda3 years ago
4 0

<em><u>Answer:</u></em>

<em><u></u></em>\boxed{r=-6}

<em><u>Step-by-step explanation:</u></em>

All~you~have~to~do~is~collect ~the ~constants~ on ~one ~side ~of ~the~ =~ and the~variables~(including~terms~with~r)~ on ~the~ other~ side.~ Then, manipulate~it,~ so~ that~ you~ only~ have~ a ~single ~r~on~ its~ own~ on~ one~ side and ~everything~ else~ on~ the ~other.

<em><u>Applying the method :</u></em>

→ Subtract ~6r~from~ both ~sides:<em><u /></em>

<em><u /></em>6r-6r+7~~=~~13+7r-6r

         0+7~~=~~13+r<em><u /></em>

<em><u /></em>

→ Subtract~13~from~both~sides:

7-13~~=~~13-13+r

     -6~~=~~0+r

          r=-6

<em><u /></em>

<em><u /></em>

<em><u>The short cut approach for add or subtract is:</u></em>

<em><u /></em>Move~ something~ to~ the~ other ~side ~of~ the~ ``=" ~and ~change~its~ sign.<em><u /></em>

<em><u /></em>

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Only for c. n = 45 d. n = 55 e. n = 110 f. n = 440 we can ensure that we can apply the normal approximation for the sample mean

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Step-by-step explanation:

For this case we know that for a random variable X we have the following parameters given:

\mu = 102, \sigma =10

Since we can't assume that the distribution of X is the normal then we need to apply the central limit theorem in order to approximate the \bar X with a normal distribution. And we need to check if n>30 since we need a sample size large as possible to assume this.

\bar X \sim N (\mu ,\frac{\sigma}{\sqrt{n}} )

Based on this rule we can conclude:

a. n = 14 b. n = 19 c. n = 45 d. n = 55 e. n = 110 f. n = 440

Only for c. n = 45 d. n = 55 e. n = 110 f. n = 440 we can ensure that we can apply the normal approximation for the sample mean

for n=14 or n =19 since the sample size is <30 we don't have enough evidence to conclude that the sample mean is normally distributed

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