Answer:
Do I really need it?
Can I find it for a cheaper price?
Can I replace it with something else that I already have or that has a lower price?
Step-by-step explanation:
When you are trying to build up your savings you need to to spent money only on the necessary things to be able to save more money every time you get paid. So, according to this, the three questions that Aria should ask herself before making a purchase are:
-Do I really need it? To make sure that it is something that you really require and not something that you just want to buy but it is not going to cover a need.
-Can I find it for a cheaper price? Maybe you can buy it in another place where the product has a lower price or you can get a cheaper brand.
-Can I replace it with something else that I already have or that has a lower price? Maybe you already have a product that does the same thing or you can buy one that replaces the function this one does and that is cheaper.
Answer:
False
Step-by-step explanation:
2cos^2(4x) - 1 = cos(8x)
cos(8 *pi/12) = cos(2pi/3) = cos(pi - pi/3) = - cos(pi/3)= 1/2, which is not 0
cos(8x) = 0
==> 8x = 0 or pi
==> x = 0 or pi/8
Answer:
ANSWER
n < - 3 \: or \: n > - 2n<−3orn>−2
EXPLANATION
The given inequality is,
|2n + 5| \: > \: 1∣2n+5∣>1
By the definition of absolute value,
- (2n + 5) \: > \: 1 \: or \: (2n + 5) \: > \: 1−(2n+5)>1or(2n+5)>1
We divide through by negative 1, in the first part of the inequality and reverse the sign to get,
2n + 5 \: < \: - 1 \: or \: (2n + 5) \: > \: 12n+5<−1or(2n+5)>1
We simplify now to get,
2n \: < \: - 1 - 5 \: or \: 2n \: > \: 1 - 52n<−1−5or2n>1−5
2n \: < \: - 6 \: or \: 2n \: > \: - 42n<−6or2n>−4
Divide through by 2 to obtain,
n \: < \: - 3 \: or \: n \: > \: - 2n<−3orn>−2
Since each trial has the same probability of success,
Let, <span><span><span>Xi</span>=1</span></span> if the <span><span>i<span>th</span></span></span> trial is a success (<span>0</span> otherwise). Then, <span><span>X=<span>∑3<span>i=1</span></span><span>Xi</span></span><span>X=<span>∑<span>i=1</span>3</span><span>Xi</span></span></span>,
and <span><span>E[X]=E[<span>∑3<span>i=1</span></span><span>Xi</span>]=<span>∑3<span>i=1</span></span>E[<span>Xi</span>]=<span>∑3<span>i=1</span></span>p=3p=1.8</span><span>E[X]=E[<span>∑<span>i=1</span>3</span><span>Xi</span>]=<span>∑<span>i=1</span>3</span>E[<span>Xi</span>]=<span>∑<span>i=1</span>3</span>p=3p=1.8</span></span>
So, <span><span>p=0.6</span><span>p=0.6</span></span>, and <span><span>P{X=3}=<span>0.63</span></span><span>P{X=3}=<span>0.63</span></span></span>
I thought what I did was sound, but the textbook says the answer to (a) is <span>0.60.6</span> and (b) is <span>00</span>.
Their reasoning (for (a)) is as follows:
