Answer:
x = 4
Move variables to the right and constants to the left. Then divide.
Step-by-step explanation:
5x + 10 = 7.5x (Given)
-2.5x = -10 (subtraction property of equality)
x = 4 (division property of equality)
Y= -3 - 2x because all you have to do is move the 2x over to have an x and a y on different sides, and in order to do that you do the opposite so since it is a positive 2, you subtract it and make it negative
First, you want to solve for the equation for this problem, would be:
0.05N + 0.10D = 20.50
While N = The amount of nickels, and D = The amount of dimes.
Since N = 164, and D = 123. It would add up to 287 coins. 164 + 123 = 287.
Now that you have the number for those two variables, solve the equation for when N = 164, and D = 123.
0.05N + 0.10D = 20.50
<span>0.05(164) + 0.10(123) = 20.50
</span><span>8.2 + 12.3 = 20.50
</span>20.50 = 20.50
So, there is 164 nickels, and 123 dimes for your answer.
<em>I hope this helps! </em>
<em>~ Notorious Sovereign</em>
Step-by-step explanation:
In statistics, the empirical rule states that for a normally distributed random variable,
- 68.27% of the data lies within one standard deviation of the mean.
- 95.45% of the data lies within two standard deviations of the mean.
- 99.73% of the data lies within three standard deviations of the mean.
In mathematical notation, as shown in the figure below (for a standard normal distribution), the empirical rule is described as
![\Phi(\mu \ - \ \sigma \ \leq X \ \leq \mu \ + \ \sigma) \ = \ 0.6827 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi(\mu \ - \ 2\sigma \ \leq X \ \leq \mu \ + \ 2\sigma) \ = \ 0.9545 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi}(\mu \ - \ 3\sigma \ \leq X \ \leq \mu \ + \ 3\sigma) \ = \ 0.9973 \qquad (4 \ \text{s.f.})](https://tex.z-dn.net/?f=%5CPhi%28%5Cmu%20%5C%20-%20%5C%20%5Csigma%20%5C%20%5Cleq%20X%20%5C%20%5Cleq%20%5Cmu%20%5C%20%2B%20%5C%20%5Csigma%29%20%5C%20%3D%20%5C%200.6827%20%5Cqquad%20%284%20%5C%20%5Ctext%7Bs.f.%7D%29%20%5C%5C%20%5C%5C%20%5C%5C%20%5CPhi%28%5Cmu%20%5C%20-%20%5C%202%5Csigma%20%5C%20%5Cleq%20X%20%5C%20%5Cleq%20%5Cmu%20%5C%20%2B%20%5C%202%5Csigma%29%20%5C%20%3D%20%5C%200.9545%20%5Cqquad%20%284%20%5C%20%5Ctext%7Bs.f.%7D%29%20%5C%5C%20%5C%5C%20%5C%5C%20%5CPhi%7D%28%5Cmu%20%5C%20-%20%5C%203%5Csigma%20%5C%20%5Cleq%20X%20%5C%20%5Cleq%20%5Cmu%20%5C%20%2B%20%5C%203%5Csigma%29%20%5C%20%3D%20%5C%200.9973%20%5Cqquad%20%284%20%5C%20%5Ctext%7Bs.f.%7D%29)
where the symbol
(the uppercase greek alphabet phi) is the cumulative density function of the normal distribution,
is the mean and
is the standard deviation of the normal distribution defined as
.
According to the empirical rule stated above, the interval that contains the prices of 99.7% of college textbooks for a normal distribution
,
![\Phi(113 \ - \ 3 \ \times \ 12 \ \leq \ X \ \leq \ 113 \ + \ 3 \ \times \ 12) \ = \ 0.9973 \\ \\ \\ \-\hspace{1.75cm} \Phi(113 \ - \ 36 \ \leq \ X \ \leq \ 113 \ + \ 36) \ = \ 0.9973 \\ \\ \\ \-\hspace{3.95cm} \Phi(77 \ \leq \ X \ \leq \ 149) \ = \ 0.9973](https://tex.z-dn.net/?f=%5CPhi%28113%20%5C%20-%20%5C%203%20%5C%20%5Ctimes%20%5C%2012%20%5C%20%5Cleq%20%5C%20X%20%5C%20%5Cleq%20%5C%20113%20%5C%20%2B%20%5C%203%20%5C%20%5Ctimes%20%5C%2012%29%20%5C%20%3D%20%5C%200.9973%20%5C%5C%20%5C%5C%20%5C%5C%20%5C-%5Chspace%7B1.75cm%7D%20%5CPhi%28113%20%5C%20-%20%5C%2036%20%5C%20%5Cleq%20%5C%20X%20%5C%20%5Cleq%20%5C%20113%20%5C%20%2B%20%5C%2036%29%20%5C%20%3D%20%5C%200.9973%20%5C%5C%20%5C%5C%20%5C%5C%20%5C-%5Chspace%7B3.95cm%7D%20%5CPhi%2877%20%5C%20%5Cleq%20%5C%20X%20%5C%20%5Cleq%20%5C%20149%29%20%5C%20%3D%20%5C%200.9973)
Therefore, the price of 99.7% of college textbooks falls inclusively between $77 and $149.