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Makovka662 [10]
3 years ago
7

Here are the scores of students on a history test. 55, 56, 57, 58, 58, 64, 65, 66, 69, 75, 77, 78, 81, 81, 85, 86, 91. Create a

box and whisker plot

Mathematics
1 answer:
Anarel [89]3 years ago
6 0
The first thing you need to do is get the median of the whole set.

The median is the center so since there are 17 numbers the center is 69 and 69 is Q2

Next we need to find Q1 Q3 and Q4 Also, are whiskers are below.
Whisker minimum = 55
Whisker maximum = 91

Now to find Q1 we have to get the median of the first half of the data.
We need to get the median of the following data set.
55, 56, 57, 58, 58, 64, 65, 66

(58+58) / 2 = 58
Q1 = 58

Now we need to find Q3. To do that we get the median of the second half of the data.
75, 77, 78, 81, 81, 85, 86, 91. 
(81 + 81) / 2 = 81
Q3 = 81

Now we have all our quartiles and whiskers, which are below, it is time to create our box and whisker plot. Note, create a horizontal or vertical number line with appropriate data. Also the box part is from Q1 to Q3 and your whiskers goes from your minimum to your box and from your maximum to your box.

Whisker Minimum = 55
Q1 = 58
Q2 = 69
Q3 = 81
Whisker Maximum = 91

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Correctly identify the property shown.
sleet_krkn [62]
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PLEASE HELP! Determine the slope and the y-intercept of the table.
natita [175]

Let's take the first 2 x and y values and use them. First, you need to fill in your formula: y2-y1/x2-x1. It should look like this: 5-2/4-2. Then solve and you should get 3/2. You can't simplify it. So, this is your slope.

Next, you need to put this is into another formula. The formula is y-y1=m(x-x1). Take the first coordinate pair and plug it in for x1 and y1. Plug in your slope for m. When it's filled in, you should have y-2+3/2(x-1). It's hard to explain how to solve it so i'll just write it out.

y-2=3/2(x-2)

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y=3/2x-1

So, we have our equation in slope intercept form so that we have our slope and our y-intercept.

Y-intercept: -1

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Please put this as brainliest it took awhile and I hope this helped :) reply with questions if you still need help or have questions about the steps

3 0
3 years ago
For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.
nordsb [41]

Answer:

<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>

By De morgan's law

(A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq  P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq  1-P(A)+1-P(B)\\\\-P(A\cap B)\leq  1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1

which is Bonferroni’s inequality

<h3>Result 1: P (Ac) = 1 − P(A)</h3>

Proof

If S is universal set then

A\cup A^{c}=S\\\\P(A\cup A^{c})=P(S)\\\\P(A)+P(A^{c})=1\\\\P(A^{c})=1-P(A)

<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>

Proof:

If S is a universal set then:

A\cup(B\cap A^{c})=(A\cup B) \cap (A\cup A^{c})\\=(A\cup B) \cap S\\A\cup(B\cap A^{c})=(A\cup B)

Which show A∪B can be expressed as union of two disjoint sets.

If A and (B∩Ac) are two disjoint sets then

P(A\cup B) =P(A) + P(B\cap A^{c})---(1)\\

B can be  expressed as:

B=B\cap(A\cup A^{c})\\

If B is intersection of two disjoint sets then

P(B)=P(B\cap(A)+P(B\cup A^{c})\\P(B\cup A^{c}=P(B)-P(B\cap A)

Then (1) becomes

P(A\cup B) =P(A) +P(B)-P(A\cap B)\\

<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>

Proof:

If A and B are two disjoint sets then

A=A\cap(B\cup B^{c})\\A=(A\cap B) \cup (A\cap B^{c})\\P(A)=P(A\cap B) + P(A\cap B^{c})\\

<h3>Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) </h3>

Proof:

If B is subset of A then all elements of B lie in A so A ∩ B =B

A =(A \cap B)\cup (A\cap B^{c}) = B \cup ( A\cap B^{c})

where A and A ∩ Bc  are disjoint.

P(A)=P(B\cup ( A\cap B^{c}))\\\\P(A)=P(B)+P( A\cap B^{c})

From axiom P(E)≥0

P( A\cap B^{c})\geq 0\\\\P(A)-P(B)=P( A\cap B^{c})\\P(A)=P(B)+P(A\cap B^{c})\geq P(B)

Therefore,

P(A)≥P(B)

8 0
3 years ago
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