Answer: The lowest value: 100 and highest value: - 150 .
If we were to build the box plot for this data, the box would stretch between
and
.
Step-by-step explanation:
We know that the box-plot is the graphical way to represent the five -number summary (Minimum value , First quartile
, Median , Third Quartile
, Maximum value).
Where , the box streches between the first quartile
and the third quartile
.
Given : The cost of taking your pet aboard the air flight with you in the continental US varies according to the airlines.
The five number summary for prices based on a sample of major US airlines was:
Min = 60,

Median = 110

Max = 150
If we were to build the box plot for this data, the box would stretch between
and
.
Hence, the lowest value: 100 and highest value: 125 .
A meter is 100 cm. A ball of string has 10m of string, and you multiply 10m by 100cm, you get 1,000cm. Then you multiply 40cm by 2, and you get 80cm, or two lengths of string. Then you multiply 35, which is half of 70, by 80, and that is 2,800 cm.
2,800÷40cm=70
And 100cm=1m.
The teacher will therefore need 28 balls of string.
Answer:
The answer is below
Step-by-step explanation:
A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?
Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

a) P(x > 5) = 
b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.
That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757
c) Let b be the amount of raw sugar should be stocked for the plant each day.
P(x > a) = 
But P(x > a) = 0.05
Therefore:
![e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98](https://tex.z-dn.net/?f=e%5E%7B-0.25a%7D%3D0.05%5C%5Cln%5Be%5E%7B-0.25a%7D%5D%3Dln%280.05%29%5C%5C-0.25a%3D-2.9957%5C%5Ca%3D11.98)
a ≅ 12