36 questions-----3 minutes
x questiones-----1 minutes
x=(1*36)/3
x=36/3
x=12 questions
B.
Answer:
The 95% confidence interval estimate of the true population proportion of U.S. employers that were likely to require higher employee contributions for health care coverage is 0.52 +/- 0.0370
= (0.4830, 0.5570)
The margin of error M.E = 0.0370
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
p+/-z√(p(1-p)/n)
p+/-M.E
Given that;
M.E = margin of error
Proportion p = 52% = 0.52
Number of samples n = 700
Confidence interval = 95%
z value (at 95% confidence) = 1.96
Substituting the values we have;
0.52 +/- 1.96√(0.52(1-0.52)/700)
0.52 +/- 1.96(0.0189)
0.52 +/- 0.0370
( 0.4830, 0.5570)
The 95% confidence interval estimate of the true population proportion of U.S. employers that were likely to require higher employee contributions for health care coverage is 0.52 +/- 0.0370
= (0.4830, 0.5570)
The margin of error M.E = 0.0370
This is the concept of application of quadratic equations; To get the number of dimes and quarters we proceed as follows;
suppose there are x dimes and y quarters;
x+y=95.......i
but we know;
$0.1=1 dimes
$0.25=1 quarters
thus
0.1x+0.25y=23.45.....ii
solving equation i and ii by substitution we shall have:
from i;
x=95-y
thus substituting the value of x in equation ii we get
0.1(95-y)+0.25y=23.45
9.5-0.1y+0.25y=23.45
collecting like terms we get:
0.15y=13.95
dividing both sides by 0.15 we get;
y=13.95/0.15=93
x=95-93=2
therefore we conclude that there were 2 dimes and 93 quarters
7.8 = 0.65x
x = 100/65 * 78/10
x = 10/5 * 6
x = 12
