To solve this problem, you have to simplify your answer.
You may also look online to solve your problem.
After simplifying your question, The answer to your question is:
x=9/7.
I hope this helps. This answer is varified.
The answer is 70 because if you multiply the money she spends by the number of days you get 15. Then subtract 15 from 85 and you will get 70
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Answer:
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
Step-by-step explanation:
We formulate null and alternate hypotheses are
H0 : u1 < u2 against Ha: u1 ≥ u 2
Where u1 is the group tested after they were awake for 24 hours.
The Significance level alpha is chosen to be ∝ = 0.05
The critical region t ≥ t (0.05, 13) = 1.77
Degrees of freedom is calculated df = υ= n1+n2- 2= 5+10-2= 13
Here the difference between the sample means is x`1- x`2= 35-24= 11
The pooled estimate for the common variance σ² is
Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]
= 1/13 [ 120²+360²]
Sp = 105.25
The test statistic is
t = (x`1- x` ) /. Sp √1/n1 + 1/n2
t= 11/ 105.25 √1/5+ 1/10
t= 11/57.65
t= 0.1908
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
Answer:
a) The probability that the airline will lose no bags next monday is 0.1108
b) The probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227
c) I would recommend taking a Poisson model with mean 4.4 instead of a Poisson model with mean 2.2
Step-by-step explanation:
The probability mass function of X, for which we denote the amount of bags lost next monday is given by this formula
![P(X=k) = \frac{e^{-2.2} * {2.2}^k }{k!}](https://tex.z-dn.net/?f=%20P%28X%3Dk%29%20%3D%20%5Cfrac%7Be%5E%7B-2.2%7D%20%2A%20%7B2.2%7D%5Ek%20%7D%7Bk%21%7D%20)
a)
![P(X=0) = \frac{e^{-2.2} * {2.2}^0 }{0!} = 0.1108](https://tex.z-dn.net/?f=%20%20P%28X%3D0%29%20%3D%20%5Cfrac%7Be%5E%7B-2.2%7D%20%2A%20%7B2.2%7D%5E0%20%7D%7B0%21%7D%20%3D%200.1108%20)
The probability that the airline will lose no bags next monday is 0.1108.
b) Note that
. And
![P(X=0)+P(X=1)+P(X=2) = e^{-2.2} * (1 + 2.2 + 2.2^2/2) = 0.6227](https://tex.z-dn.net/?f=P%28X%3D0%29%2BP%28X%3D1%29%2BP%28X%3D2%29%20%3D%20e%5E%7B-2.2%7D%20%2A%20%281%20%2B%202.2%20%2B%202.2%5E2%2F2%29%20%3D%200.6227)
Therefore, the probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227.
c) If the double of flights are taken, then you at least should expect to loose a similar proportion in bags, because you will have more chances for a bag to be lost. WIth this in mind, we can correctly think that the average amount of bags that will be lost each day will double. Thus, i would double the mean of the Poisson model, in other words, i would take a Poisson model with mean 4.4, instead of 2.2.
We know that <span>Lola walks 4/10 mile to her friends house. then she walks 5/10 mile to the store.
To solve the problem we have to use addition.
4/10 + 5/10 = 9/10
Lola walks 9/10 miles. </span>