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babunello [35]
3 years ago
12

Which could be the missing data item for the given set of data if the median of the complete data set is 15? 11, 23, 12, 18, 11,

10, 19, 15, 19, 21, 13, 17, 24, 14
Mathematics
1 answer:
Kipish [7]3 years ago
4 0
Any number below 15, because the median (or very middle number) of this data set is between 15 and 17, so to make the median 15 one would need to add a number below 15 to counter  17, so that there would be an equal amount of numbers on either side of 15.
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Find the average of the following numbers −30 and 25
alina1380 [7]

Answer: -2.5

Step-by-step explanation:

The average can be found by adding up the numbers then dividing by how many numbers there are in this case there are two numbers.

-30+25=-5

-5/2 = -2.5

7 0
3 years ago
Type an expression using words to represent each of the given algebraic expressions. Enter your answer in the box
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Step-by-step explanation:

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2 years ago
I have homework for tomorrow can someone help
Tema [17]

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x = 42 degrees.

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3 0
2 years ago
Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

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E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
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