21 students because if 6 represents 30% u add 6 two more time which equals 18 and 90% then u add 3 more because it is half of 6. when u add 3 more u will get 21 and 100%
A. The point estimate of μ1 − μ2 is calculated using the value of x1 - x2, therefore:
μ1 − μ2 = x1 – x2 =
7.82 – 5.99
μ1 − μ2 = 1.83
B. The formula for
confidence interval is given as:
Confidence interval
= (x1 –x2) ± z σ
where z is a value
taken from the standard distribution tables at 99% confidence interval, z =
2.58
and σ is calculated
using the formula:
σ = sqrt [(σ1^2 /
n1) + (σ2^2 / n2)]
σ = sqrt [(2.35^2 /
18) + (3.17^2 / 15)]
σ = 0.988297
Going back to the
confidence interval:
Confidence interval
= 1.83 ± (2.58) (0.988297)
Confidence interval
= 1.83 ± 2.55
Confidence interval
= -0.72, 4.38
Answer:

Step-by-step explanation:
- If f(x) is in th form of f(x)=g(x)-h(x) then f'(x)=g'(x) - h'(x)
- When f(x)=z(g(x)) then f'(x)= z'(g(x))g'(x) (called as chain rule)
<u>using these information</u>:
g(x)=ln2x then g'(x)=
h(x)=In(3x - 1) then h'(x)=![\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}](https://tex.z-dn.net/?f=%5Cfrac%7B%283x-1%29%27%7D%7B3x-1%7D%20%3D%5Cfrac%7B3%7D%7B3x-1%7D%3C%2Fp%3E%3Cp%3Ef%27%28x%29%3Dg%27%28x%29%20-%20h%27%28x%29%20%3D%5Btex%5D%5Cfrac%7B1%7D%7Bx%7D%20-%20%5Cfrac%7B3%7D%7B3x-1%7D%20%3D%5Cfrac%7B-1%7D%7B3x%5E2-x%7D)
Using the arrangements formula, considering the given digits, it is found that Edward writes 6 numbers.
<h3>What is the arrangements formula?</h3>
The number of possible arrangements of n elements is given by the factorial of n, that is:

In this problem, the first digit has to be 5, while the other three are arranged, hence the total number of arrangements is given by:

More can be learned about the arrangements formula at brainly.com/question/25925367
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Answer:
y = 4
Step-by-step explanation:
Since the line if parallel to the x-axis, regardless of the value of x, y will be equal to 4. In general, then, the points on the line are (x,4), where x is any value, including 1.