What values of b satisfy 3(2b+3)^2 = 36
we have
3(2b+3)^2 = 36
Divide both sides by 3
(2b+3)^2 = 12
take the square root of both sides
( 2b+3)} =(+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3
b1=\frac{\sqrt{12}}{2} -\frac{3}{2}
b1=\sqrt{3} -\frac{3}{2}
b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
therefore
the answer is
the values of b are
b1=\sqrt{3} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
A sideways opening parabola is in the form

, so we know from the process of elimination that it will either be b or c. Next we have to realize that if the parabola opens to the left it is a negative parabola, just like if a parabola opens upside down it is a negative parabola. So the one that has the negative out front is b.
<span>m∠SYD = </span>106.02°
..............................................................
Answer:
The correct answer is
(0.0128, 0.0532)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence interval
, we have the following confidence interval of proportions.

In which
Z is the zscore that has a pvalue of 
For this problem, we have that:
In a random sample of 300 circuits, 10 are defective. This means that
and 
Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.
So
= 0.05, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The correct answer is
(0.0128, 0.0532)