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3 years ago

I've never even seen a box and whiskers plot before!! Should be pretty straight forward though. Thanks!

Mathematics

2 answers:

dybincka [34]3 years ago

8 0

So a box and whiskers plot is relatively simple if you know how to do it.

The left whisker ending, which is at 35, signifies the minimum.

The right whisker ending, which is at 99, signifies the maximum.

The middle line, at 68, is the median.

And the edge of the box on the left, at 50, is the First Quartile. (If you don't know what that is, basically the marker in between the first quarter and the second quarter)

And the edge of the box on the right is the Third Quartile, (Same thing except in between the third quarter and fourth)

So since the question is asking how many students scored better than an 85, and we know that 85 is the third quartile here, we know that 1/4 of the students scored better than an 85.

(So basically each section is 1/4 of the data set)

And 1/4 of 24 is 6.

So 6 students scored better than an 85.

alisha [4.7K]3 years ago

7 0

For a box and whispers plot, each section represents 25% of the data. Therefore, the area from 85-99 represents 25% of the data.

25% of 24 students= 6 students
Final answer: 6 students

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Kelsey graphed the equation y = 3x + 1 as shown below.

USPshnik [31]

Answer:

D. Kelsey graphed the slope as the y-intercept and the y-intercept as the slope.

Step-by-step explanation:

The equation to graph was : y= 3x + 1 where the slope is 3 and the y-intercept is 1

However, Kelsey graph line passes through points (-3, 0)and (0, 3).From these points the equation of the line is;

m=Δy/Δx where

m=the slope of the graph

Δy= 3-0 =3

Δx= 0--3 =3

m=3/3 = 1

The equation of the line should be;

m=Δy/Δx

1=y-3/x-0

1x= y-3

3+1x= y

y= 1x + 3 ----------where the slope is 1 and y-intercept is 3

So you can see here that in the original equation, the slope m is 3 and y-intercept is 1 as shown in the first attached graph.

While in the Kelsey's graph, the slope is 1 and the y-intercept is 3 as in the second attached graph.

Thus, the correct answer is D : Kelsey graphed the slope,3 as the y-intercept and the y-intercept,1, as the slope.

Answer choice A is incorrect because Kelsey didnot graph the y-intercept, 1 on the x-axis but graphed point (-3,0) on the x-axis.

Answer choice B is incorrect because on Kelsey's equation the y-intercept is 3 and not -3. i.e. y=1x + 3

Answer choice C is incorrect because Kelsey graphed the slope as up 1 right 1 which is okay per the equation , y=1x+3.However, this incorrect because the correct graph has a slope of 3.

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Given 2x^2=-8, x is A.-2 B. 2 C. 2i or -2i D. 4i or -4i

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Answer:

2i or -2i

Step-by-step explanation:

2x² = -8 First, clear x.

x² = -8/2 Solve

x² = -4 Now, eliminate the square of x by solving the square root of -4.

x = √-4

x = 2i or -2i Is an imaginary number.

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3 years ago

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ZanzabumX [31]

Answer:

1/4

Step-by-step explanation:

The maximum sum of the two cubes is 6 + 6 = 12

The multiples of 4 until 12 are: 4, 8 and 12.

each cube has 6 possible values, so two cubes have a total of 6 * 6 = 36 possible results

We can have a sum of 4 with the pairs (1,3), (2,2), (3,1).

We can have a sum of 8 with the pairs (2,6), (3,5), (4,4), (5,3), (6,2)

We can have a sum of 12 with the pair (6,6)

So we have a total of 9 pairs that satisfy our condition over a total of 36 possibilities, so the probability is 9 / 36 = 1/4

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3 years ago

Which of the following is not a property of a chi-square distribution?

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Answer:

c) Is not a property (hence (d) is not either)

Step-by-step explanation:

Remember that the chi square distribution with k degrees of freedom has this formula

$\chi_k^2 = \matchal{N}_1^2 + \matchal{N}_2^2 + ... + \, \matchal{N}_{k-1}^2 + \matchal{N}_k^2$

Where N₁ , N₂m .... $N_k$ are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.

Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true

The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.

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3 years ago

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Answer:

3 boxes?

Step-by-step explanation:

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