Answer:
Step-by-step explanation:
a )
sample mean = sum total of given data / no of data
= 415.35 / 20 = 20.76
To calculate the median we arrange the data in ascending order and take the average of 10 th and 11 th term .
= 20.50 + 20.72 / 2
= 20.61
b ) To calculate the 10% trimmed mean , we neglect the largest 10% and smallest 10 % data and then calculate the mean . Here we neglect the first two smallest and last two greatest
(18.92 + 19.25 ..... + 22.43 + 22.85) / 16
= 20.74
c )
We can easily plot the data on number line from 17 to 24
d )
Maximum value of data set = 23.71 and minimum value is 18.04
mean is 20.76 , median is 20.61 and trimmed mean is 20.74
They are between maximum and minimum values of given data . Hence there is no outliers .
Angles RLN and MLK would be vertical angles.
Right. Vertical angles are formed when their
sides share the same lines. RL shares the same line with LM and NL shares the
same line with LK (see the attached diagram), so that means both angles form a vertical
pair.
Angles RLN and MLN would be vertical angles.
Wrong. They are linear pairs, because they
are adjacent and supplementary. Adjacent angles share a side – in this case,
LN. Supplementary angles sum 180°, which you can see is right because the other
sides (ML and RL) are in the same line. RLN and MLN sum the same as the size of
RLM, which is a line, so it’s 180°.
<span>
Angles RLN and KLM would be a linear pair. </span>
Wrong. They would be a vertical pair (see
definition of vertical pair in the first option). RL is opposed to LM and LN is
opposed to KL.
Angles RLN and KLN would be a linear pair.
Wrong. KLN is actually a line, so it’s actually
180°, so it can’t be a linear pair with KLN. Linear pairs sum 180°, which is
impossible because KLN itself is already 180°, so any sum will throw a higher
number.
A randomly selected car with no 4 wheel drive....so there are 40 cars with no 4 wheel drive
cars with 3rd row seats...that do not have 4 wheel drive...there are 12
so the probability is : 12/40 = 0.3 <==
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