The area of a circle is A = πr^2. We let A1 And A2 the areas of the circles and r1 and r2 the radius of each, respectivley.
A1 + A2 = 80π
Substitute the formula for the area,
π(r1)^2 + π (r2)^2 = 80π
From the statement, we know that r2=2(r1).
<span>π(r1)^2 + π (2 x r1)^2 = 80π
</span>We can cancel π, we will have
5 x (r1)^2 = 80
Thus,
r1 = 4 and r2 = 8
Answer: x = -7/2
Step-by-step explanation:
10x+ 30 = 20 +8x+ 3
10x+ 30=23+8x
10x-8x = 23-30
2x=-7
( a helpful app to help with these type of problems is photomath)
Answer:
a) 95% of the widget weights lie between 29 and 57 ounces.
b) What percentage of the widget weights lie between 12 and 57 ounces? about 97.5%
c) What percentage of the widget weights lie above 30? about 97.5%
Step-by-step explanation:
The empirical rule for a mean of 43 and a standard deviation of 7 is shown below.
a) 29 represents two standard deviations below the mean, and 57 represents two standard deviations above the mean, so, 95% of the widget weights lie between 29 and 57 ounces.
b) 22 represents three standard deviations below the mean, and the percentage of the widget weights below 22 is only 0.15%. We can say that the percentage of widget weights below 12 is about 0. Equivalently we can say that the percentage of widget weights between 12 an 43 is about 50% and the percentage of widget weights between 43 and 57 is 47.5%. Therefore, the percentage of the widget weights that lie between 12 and 57 ounces is about 97.5%
c) The percentage of widget weights that lie above 29 is 47.5% + 50% = 97.5%. We can consider that the percentage of the widget weights that lie above 30 is about 97.5%