Answer:
B
Step-by-step explanation:
you can figure this out by replace the x and y with any one of the x, y positions and if the answer is 1 it's correct.
I tried 0^2/16+3^2/9 and I got 1.
I tried this with the first two but they didn't work.
Answer:
5/6
Step-by-step explanation:
I hope this helps!
<h2><u>
Answer with explanation</u>
:</h2>
Let 
be the distance traveled by deluxe tire .
As per given , we have
Null hypothesis : 
Alternative hypothesis : 
Since 
is left-tailed and population standard deviation is known, thus we should perform left-tailed z-test.
Test statistic : 
where, n= sample size
= sample mean
= Population mean
=sample standard deviation
For 
, we have
By using z-value table,
P-value for left tailed test : P(z≤-2.23)=1-P(z<2.23) [∵P(Z≤-z)=1-P(Z≤z)]
=1-0.9871=0.0129
Decision : Since p value (0.0129) < significance level (0.05), so we reject the null hypothesis .
[We reject the null hypothesis when p-value is less than the significance level .]
Conclusion : We do not have enough evidence at 0.05 significance level to support the claim that t its deluxe tire averages at least 50,000 miles before it needs to be replaced.
Answer:
0.34134
Step-by-step explanation:
In other to solve for this question, we would be using the z score formula
z = (x - μ) / σ
x = raw score
μ = mean
σ = Standard deviation
We are told in the question to find the probability that a worker selected at random makes between $350 and $400
let x1 = 350 and x2= 400 with the mean μ = 400 and standard deviation σ = $50.
z1 = (x1 - μ) / σ = (350-400) / 50 = -1
z2 = (x2 - μ) / σ = (400 - 400) / 50 = (0/50) = 0
From tables, P(z <= -1) = 0.15866
P(z <= 0) = 0.5
Then, the probability would give us, P(-1 ≤ z ≤ 0) =0.5 - 0.15866 =
0.34134
Hence, The probability that a worker selected at random makes between $350 and $400 = 0.34134
