Answer:
x= 1/22 and y=-1/4 (I'm assuming that the equal signs are the only separators of the equation sequence, since there is no indicator that the so-called "sentences" are actually apart.
Please, post the instructions along with your question, and if possible share the question in symbolic form, not in words.
Do you mean Question #5? By (2) x cube, do you mean 2x^3?
I strongly suggest that you use lots of parentheses ( ) to show how your numbers are grouped, and not to use " x " to denote multiplication (use " * " for that, please.
If only you'll clear this up, I'd be happy to help.
I will assume that your post is 2x^3 - 3(9x-5)^2.
Then 2x^3 - 3(81x - 90x + 25). Does this have any resemblance to what you wanted me to see in your post?
We can solve this problem by referring to the standard
probability distribution tables for z.
We are required to find for the number of samples given the
proportion (P = 5% = 0.05) and confidence level of 95%. This would give a value
of z equivalent to:
z = 1.96
Since the problem states that it should be within the true
proportion then p = 0.5
Now we can find for the sample size using the formula:
n = (z^2) p q /E^2
where,
<span> p = 0.5</span>
q = 1 – p = 0.5
E = estimate of 5% = 0.05
Substituting:
n = (1.96^2) 0.5 * 0.5 / 0.05^2
n = 384.16
<span>Around 385students are required.</span>
Answer: The total number of logs in the pile is 6.
Step-by-step explanation: Given that a stack of logs has 32 logs on the bottom layer. Each subsequent layer has 6 fewer logs than the previous layer and the top layer has two logs.
We are to find the total number of logs in the pile.
Let n represents the total number of logs in the pile.
Since each subsequent layer has 6 fewer logs then the previous layer, so the number of logs in each layer will become an ARITHMETIC sequence with
first term, a = 32 and common difference, d = -6.
We know that
the n-th term of an arithmetic sequence with first term a and common difference d is
Since there are n logs in the pile, so we get
Thus, the total number of logs in the pile is 6.
