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GalinKa [24]
3 years ago
10

The box and whiskers plot represents the results of the same math test being given to two different classes. According to the bo

x
plot, which statement is true?
A) Both classes have a range of 45
B) Both classes have a median of 85
C) Both classes have a median of 100
D) Approximately 1/4 of second period students scored an 85 or above​

Mathematics
2 answers:
densk [106]3 years ago
4 0

Answer: Answer A: Both classes have a range of 45

Step-by-step explanation:

above the two graphs we can see a number line ranging from 50-100

bellow that, we can see 2 box and whisker plots, one for second period and one for fourth period. the larger rectangular boxes show all the number results, the line in between them states the average result. Both lines have starting points at 55 and ending points at 100.

100-55=45   45 being the range. Since 2nd and 4th both share the same range answer A could possibly be correct

B is wrong because 2nd period has a median of 90

C is wrong because no periods have a median of 100

D is wrong because most of the students scored an 85+

conclusion: answer A is correct

creativ13 [48]3 years ago
4 0

Answer: A

Step-by-step explanation: To find the range of each class subtract 55 from 100, which is 45.

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2 years ago
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ludmilkaskok [199]

Answer:

\lambda \geq 6.63835

Step-by-step explanation:

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Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 2)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}

And replacing we have this:

P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)

And we want this probability that at least of 99%, so we can set upt the following inequality:

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x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

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f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

\lambda \geq 6.63835

4 0
3 years ago
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