Fist we find the legnth of the diagonal
a^2+b^2=c^2
3^2+4^2=c^2
9+16=c^2
25=c^2
c=5
now the other diagonal
11^2+5^2=c^2
121+25=c^2
146=c^2
sqrt both sides12.083
closest is B
Answer: y = -5
Substitute in 3 for x.
x - y = 8 Substitution
3 - y = 8 Subtract 3 from both sides
-y = 5 Divide by -1 to both sides
y = -5 Answer!
You can also check your work.
x - y = 8
3 - (-5) = 8
3 + 5 = 8
8 = 8
Mean weight of the bag of pears = u = 8 pounds
Standard deviation = s = 0.5 pounds
We have to find what percentage of bags of pears will weigh more than 8.25 pounds. This can be done using the z score.
We have to convert x = 8.25 to z scores, which will be:
z score = 0.5From the z table, the probability of z score being greater than 0.5 is 0.3085
Therefore, the probability of a bag to weigh more than 8.25 pounds is 0.3085
Thus 0.3085 or 31% (rounded to nearest integer) of bags of pears will have weight more than 8.25 pounds at the local market.
<em>x</em>/<em>r</em> + <em>x</em>/<em>w</em> + <em>x</em>/<em>t</em> = 1
<em>x</em> (1/<em>r</em> + 1/<em>w</em> + 1/<em>t</em>) = 1
<em>x</em> = 1 / (1/<em>r</em> + 1/<em>w</em> + 1/<em>t</em>)
To make the solution a bit cleaner, multiply through the numerator and denominator by the LCM of each fraction's denominator, <em>rwt</em> :
<em>x</em> = 1 / (1/<em>r</em> + 1/<em>w</em> + 1/<em>t</em>) • <em>rwt</em> / <em>rwt</em>
<em>x</em> = <em>rwt</em> / (<em>rwt</em>/<em>r</em> + <em>rwt</em>/<em>w</em> + <em>rwt</em>/<em>t</em>)
<em>x</em> = <em>rwt</em> / (<em>wt</em> + <em>rt</em> + <em>rw</em>)