The confidence interval formula is computed by:
Xbar ± Z s/ sqrt (n)
Where:
Xbar is the mean
Z is the z value
S is the standard deviation
N is the number of samples
So our given are:
90% confidence interval with a z value of 1.645
Sample size 40, 45
Mean 180, 179
Standard deviation 2, 4
So plugging that information in the data will give us a
confidence interval:
For 1:
Xbar ± Z s/ sqrt (n)
= 180 ± 1.645 (2 / sqrt (40))
= 180 ± 1.645 (0.316227766)
= 180 ± 0.520194675
= 179.48, 180.52
For 2:
Xbar ± Z s/ sqrt (n)
= 179 ± 1.645 (4 / sqrt (45))
<span>= 179 ± 1.645 (0.596284794)</span>
therefore, the answer is letter b
100%/x%=80/20
(100/x)*x=(80/20)*x - we multiply both sides of the equation by x
100=4*x - we divide both sides of the equation by (4) to get x
100/4=x
25=x
x=25
The correct question is
<span>Teresa graphs the following 3 equations: y=2x, y=x2+2, and y=2x2. She says that the graph of y=2x will eventually surpass both of the other graphs. Is Teresa correct? Why or why not?
we have that
y=2x
y=x</span>²+2
y=2x²
using a graph tool
see the attached figure
<span>We can affirm the following
</span>the three graphs present the same domain-----> the interval (-∞,∞)
The range of the graph y=2x is the interval (-∞,∞)
The range of the graphs y=x²+2 and y=2x² is the interval [0,∞)
therefore
<span>Teresa is not correct because the graph of y = 2x will not surpass the other two graphs since in the interval of [0, infinite) the three graphs present the same range</span>
Answer:
Step-by-step explanation:
The domain of a function is all the allowed x values that make the function defined. There is nothing to make this function undefined. So the domain is negative infinity to positive infinity. All those x values are in the domain. In order to show your work, just show yourself plugging in those numbers.
You get 9, 0, and 1
Answer:
When dividing rational numbers, multiply by the "right" reciprocal. In this case, the "right" reciprocal means to take the reciprocal of the fraction on the right-hand side of the division operator.