Answer:
r = 2
r = -1
Step-by-step explanation:
So first, we need to get this is order of: ax^2 + bx + c
3r^2 - 6 - 3r
3^2 - 3r - 6
the formula is: -b ± √b^2 -4ac/2a
"ax^2 + bx + c" = "3^2 - 3r - 6"
a = 3
b = -3
c = -6
-b ± √b^2 -4ac/2a
-(-3) ± √(-3)^2 -4(3)(-6)/2(3)
3 ± √9 + (-12)(-6)/6
3 ± √9 + (72)/6
3 ± √9 + 72/6
3 ± √81/6
3 ± 9/6
3 - 9/6
-6/6
-1 = r
3 + 9/6
12/6
2 = r
Formula down below --
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V
Answer:
Step-by-step explanation:
Hello!
You need to test if the true mean of the voltage of Estonian networks is 232.
You are given a sample of n= 66 industrial networks, with sample mean X[bar]= 231.7 V and standard deviation S= 2.19 V.
With the study variable X: "voltage of an Estonian industrial network" of unknown distribution, but a large enough sample, I'll apply the Central Limit Theorem and approximate the distribution of the sample mean to normal. Remember: Since the population mean is a paramenter of the normal distribution, you need to work under it.
H₀: μ = 232
H₁: μ ≠ 232
α: 0.05 (you were given no significance level, that's why I choose to use the most common one for the test)
Z= <u>X[bar] - μ </u>≈ N (0;1)
S/√n
Z= <u>X[bar] - μ </u>= <u> 231.7 - 232 </u> = -1.113
S/√n 2.19/√66
= -1.113
This test is two tailed, and so is the p-value, in other words, you have to take into account that the p-value is divided in the two tails.
Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
P(Z<-1.113) + P(Z>1.113) = P(Z<-1.113) + (1 - P(Z< 1.113)) = 0.1335 + (1 - 0.8665) = 0.267
p-value: 0.267
Comparing it to α: 0.05, the decision is to not reject the null hypothesis.
I hope it helps!