Answer: 0.00067 minutes
Step-by-step explanation: if the proportion of customers who wait more than 15 minutes is 0.01, then the time interval between each waiting customer 15/0.01 = 1500 minutes.
The distribution that defines this question is that of an exponential.
An exponential distribution is dependent on the fixed time rate at which the event is occurring (λ)
For this question of ours, λ = 1500 minutes.
The mean of an exponential distribution is given as
u = 1/ λ = 1/1500 = 0.00067 minutes.
9514 1404 393
Answer:
x = 6
Step-by-step explanation:
The sum of segments is used:
EF +FG = EG
3x +(5x +16) = 11x -2 . . . . substitute given expressions
8x +18 = 11x . . . . . . . . add 2
18 = 3x . . . . . . . . subtract 8x
6 = x . . . . . . divide by 3
The value of the variable is 6.
__
<em>Check</em>
The segment lengths are ...
EF = 3x = 3·6 = 18
FG = 5x+16 = 5·6 +16 = 46
EG = 11x -2 = 11·6 -2 = 64 = 18+46 . . . answer is correct
Answer: Choice A) mean, there are no outliers
Have a look at the image attached below. I made two dotplots for the data points. The blue points represent bakery A. The red points represent bakery B. For any bakery, the points are fairly close together. There is no point that is off on its own. So there are no outliers, making the mean a good choice for the center. If there were outliers, then the median is a better choice. The mean is greatly affected by outliers, while the median is not.
Answer:
like this?
Step-by-step explanation:
(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 9<span>5</span>
