Answer:
(115.2642, 222.7358).
Step-by-step explanation:
Given data:
type A: n_1=60, xbar_1=1827, s_1=168
type B: n_2=180, xbar_2=1658, s_2=225
n_1 = sample size 1, n_2= sample size 2
xbar_1, xbar_2 are mean life of sample 1 and 2 respectively. Similarly, s_1 and s_2 are standard deviation of 1,2.
a=0.05, |Z(0.025)|=1.96 (from the standard normal table)
So 95% CI is
(xbar_1 -xbar_2) ± Z×√[s1^2/n1 + s2^2/n2]
=(1827-1658) ± 1.96×sqrt(168^2/60 + 225^2/180)
= (115.2642, 222.7358).
The data for resort A shows more consistency because a larger interquartile range such as the one for resort B, shows more variation. This means that the snowfall for resort A is more likely to be close to the median.
Just did this on edg. :)
<em><u>An inequality that shows the distance Johnathan could of ran any day this week is:</u></em>

<em><u>Solution:</u></em>
Let "x" be the distance Johnathan can run any day of this week
Given that,
Johnathan ran 5 days this week. The most he ran in one day was 3.5 miles
Therefore,
Number of days ran = 5
The most he ran in 1 day = 3.5 miles
Thus, the maximum distance he ran in a week is given as:

The maximum distance he ran in a week is 17.5 miles
If we let x be the distance he can run any day of this week then, we get a inequality as:

If we let y be the total distance he can travel in a week then, we may express it as,
