Answer:
The probability that 10 adults selected at random from the town all have health insurance is 0.01153.
Step-by-step explanation:
Consider the provided information.
One town, 64% of adults have health insurance.
Let p = 64% = 0.64
Therefore, q=1-0.64=0.36
We need to find the probability that 10 adults selected at random from the town all have health insurance
Use the formula: 
Here, the value of r is 10.
Substitute the respective values in the above formula.
Hence, the probability that 10 adults selected at random from the town all have health insurance is 0.01153.
Answer:
A.
This assumption is that the distribution of polyunsaturated fat % of each if these four regimes must be with equal variances as well as uniform.
B.
The null hypothesis
H0: μ1 = μ2 = μ3 = μ4
The alternate hypothesis:
H1: at least 2 means are unequal
C.
First we calculate the grand mean
= 1/54[15(42.9)+17(43.1)+8(43.7)+14(43.9)]
= (643.5 +732.7 + 349.6 + 614.6)/54
= 2340.4/54
= 43.341
Sum of squared treatment
= [15(42.9-43.341)²+17(43.1-43.341)²+8(43.7-43.341)²+14(43.9-43.341)²]
= 9.3104
Mean square of treatment
= SST/I-1
= 9.3104/4-1
= 9.3104/3
= 3.1035
Error sum of squared
= (15-1)*(1.3)² + (17-1)*(1.5)² + (8-1)*(1.2)² + (14-1)*(1.2)²
= 88.46
Error mean square
MSE = 88.46/54-4
= 1.7692
Test statistic
= 3.1035/1.7692
= 1.75
3x(4x+y)
correct me if im wrong but hope this helps
4g + 3 = 4g + 6
3 = 6
Since we know three equals six is not correct, there is no answer/result for this entire expression.
Answer:
<em>n = 20</em>
Step-by-step explanation:
[ 180° ( n - 2 ) ] / n = 162°
180 ( n - 2 ) = 162 n
180n - 360 = 162n
18n = 360
<em>n = 20</em>
