Answer:
The approximate area is A) 5.09cm²
Step-by-step explanation:
Area of a Circle = πr²
r = 1.8
Plug in our values
π(1.8cm²)
Evaluate the area of the entire circle
a = π(1.8cm²)
a = π(3.24cm)
a = 10.179cm²
Area of a sector, with the area of the circle of the sector being a = theta/360 * a
Evaluate the area of the sector
180/360*10.719cm²
0.5 * 10.179cm²
5.0895cm²
Round the value up
5.09cm²
<u>Answer:</u>
<h2>
52% were jelly beans!</h2>
<u>Explanation</u><u>:</u>
<em>Cross </em><em>multiply </em><em>the </em><em>following:</em>
<em>
</em>
<em>x </em><em>times </em><em>1</em><em>3</em><em>8</em><em> </em><em>=</em><em> </em><em>1</em><em>3</em><em>8</em><em>x</em>
<em>100 </em><em>times </em><em>7</em><em>1</em><em>.</em><em>7</em><em>6</em><em> </em><em>=</em><em> </em><em>7</em><em>,</em><em>1</em><em>7</em><em>6</em>
<em>Divide </em><em>both </em><em>sides </em><em>by </em><em>1</em><em>3</em><em>8</em><em>:</em>
<em>
</em>
<h3>
<em>x </em><em>=</em><em> </em><em>5</em><em>2</em><em>%</em></h3>
2300 cups because if 460 is 20%, multiply the 460 x 5 to get 100%
1-2. The best estimate for the population mean would be sample mean of 60 gallons. Since we know that the sample mean is the best point of estimate. Since sample size n=16 is less than 25, we use the t distribution. Assume population from normal distribution.
3. Given a=0.1, the t (0.05, df = n – 1 = 15)=1.75
4. xbar ± t*s/vn = 60 ± 1.75*20/4 = ( 51.25, 68.75)
5. Since the interval include 63, it is reasonable.
