<u><em>1/1024 is the correct answer.</em></u> First you had to apply exponent rule, and it gave us, , or
. Then you can also refine, and it gave us,
, or
. And it gave us the answer is
, or 1/1024 is the correct answer. Hope this helps! And thank you for posting your question at here on brainly, and have a great day. -Charlie
A tree is more than five times as tall as a math student. If the tree is 28 feet tall, the student can be no taller than what?
1 answer:
You might be interested in
9514 1404 393
Answer:
(x, y) ≈ (1.1642, 3.3930) and (3.4358, -0.39297)
Step-by-step explanation:
Solve the first equation for y, then substitute into the second.
5x +4/y = 7
4/y = 7 -5x
4/(7 -5x) = y
Then the second equation becomes ...
4x +x/(4/(7 -5x)) = 5
4x +x(7 -5x)/4 = 5
16x +7x -5x^2 = 20 . . . . . multiply by 4
5x^2 -23x +20 = 0 . . . . . put in standard form
We can use the quadratic formula to solve this.
x = (23±√((-23)² -4(5)(20)))/(2(5)) = (23±√129)/10
x = 2.3 ±√1.29 ≈ {1.1642, 3.4358}
y = 4/(7 -5x) = {3.3930, -0.39297}
Solutions are (x, y) ≈ (1.1642, 3.3930) and (3.4358, -0.39297).
Answer:
Step-by-step explanation:
Hello!
You have two variables of interest
X₁: failure stress of a NiCrAlZr coating after nine 1-hr cycles.
X₂: failure stress of a NiCrAlZr coating after six 1-hr cycles.
a)
To be able to estimate the difference between the means using a confidence interval, you need that both variables have a normal distribution and to determine whether or not the population variances are equal.
If the population variances are equal, σ₁²=σ₂², you can use a pooled variance t-test
If the population variances are different, σ₁²≠σ₂², you have to use Welch's t-test
Using α: 0.05
The normality test for X₁ shows a p-value of 0.7449 ⇒ You can assume it has a normal distribution.
The normality test for X₂ shows a p-value of 0.9980 ⇒ You can assume it has a normal distribution.
The F-test for variance homogeneity shows a p-value of 0.6968 (H₀:σ₁²=σ₂²) ⇒You can assume both population variances are equal.
b) and c)
You need to test if both population means are the same, the hypotheses are:
H₀: μ₁=μ₂
H₁: μ₁≠μ₂
α: 0.05
The distribution of this test is a t with 13 degrees of freedom and the test is two-tailed, so to calculate the p-value you have to do the following:
P(t₁₃≤-4.23)+P(t₁₃≥4.23)= P(t₁₃≤-4.23)+[1-P(t₁₃<4.23)]= 0.000492 + (1-0.999508)= 2*0.000492= 0.000984≅ 0.001
The p-value: 0.001 is less than α: 0.05, the decision is to reject the null hypothesis.
I hope it helps!