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jeka57 [31]
3 years ago
15

If we collect a large sample of blood platelet counts and if our sample includes a single outlier, how will that outlier appear

in a histogram?
A. The outlier will appear as a tall bar near one side of the distribution.
B. Since a histogram shows frequencies, not individual data values, the outlier will not appear. Instead, the outlier increases the frequency for its class by 1
C. The outlier will appear as the tallest bar near the center of the distribution
D. The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.
Mathematics
1 answer:
Olin [163]3 years ago
4 0

Answer:

D. The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.

Step-by-step explanation:

An histogram measures how many times each value appears in the set we are studying. That is, it is a frequency measure.

Suppose we have the following set:

S = {1,1,1,1,1,1, 2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,5,100}

1 appears 6 times. That means that when the X axis is 1, the y axis is 6.

2 appears 8 times. The means that when the X axis is 2, the y axis is 8.

...

100 appears 1 time. This means that when the X axis is 100, the y axis is 1. The X is the outlier, and it is quite far from the other values.

So the correct answer is:

D. The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.

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What are the modes of these set of data What are the modes of the following 3, 13, 6, 8, 10, 5, 6​
Tanya [424]

Answer:

Mode = 6

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8 0
3 years ago
A rubber ball dropped on a hard surface takes a sequence of bounces, each one 1/6 as high as the preceding one. If this ball is
Brums [2.3K]

Answer:

\frac{259}{54}\text{ or }4.8\text{feet}

Step-by-step explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one \frac{1}{6} as high as the preceding one.

TO FIND: If this ball is dropped from a height of 12 feet, how far will it have traveled when it hits the surface the fifth time.

SOLUTION:

once the ball is dropped on hard surface it bounces \frac{1}{6} of preceding one and comes down the same distance.

When the ball is dropped from 12 feet height

after first hit =12\times\frac{1}{6}\text{ up}+12\times\frac{1}{6}\text{ down}=4

new height =12\times\frac{1}{6}=2\text{feet}

after second hit =2\times\frac{1}{6}\text{ up}+2\times\frac{1}{6}\text{ down}=\frac{2}{3}

new height =2\times\frac{1}{6}=\frac{1}{3}\text{ feet}

after third hit  =\frac{1}{3}\times\frac{1}{6}\text{ up}+\frac{1}{3}\times\frac{1}{6}\text{ down}=\frac{1}{9}

new height =\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}\text{ feet}

after fourth hit =\frac{1}{18}\times\frac{1}{6}\text{ up}+\frac{1}{18}\times\frac{1}{6}\text{ down}=\frac{1}{54}

adding all distance =4+\frac{2}{3}+\frac{1}{9}+\frac{1}{54}

                                 =\frac{259}{54} feet

Hence the ball will travel   \frac{259}{54} feet before it hits the surface fifth time.

                                 

4 0
4 years ago
How many triangles does a=6 b=10 A=33° create?
GaryK [48]

Answer:

2 triangles are possible.

Step-by-step explanation:

Given

a=6

b=10

\angleA=33°

To find:

Number of triangles possible ?

Solution:

First of all, let us use the <em>sine rule</em>:

As per Sine Rule:

\dfrac{a}{sinA}=\dfrac{b}{sinB}

And let us find the angle B.

\dfrac{6}{sin33}=\dfrac{10}{sinB}\\sinB = \dfrac{10}{6}\times sin33\\B =sin^{-1}(1.67 \times 0.545)\\B =sin^{-1}(0.9095) =65.44^\circ

This value is in the 1st quadrant i.e. acute angle.

One more value for B is possible in the 2nd quadrant i.e. obtuse angle which is: 180 - 65.44 = 114.56^\circ

For the value of \angle B = 65.44^\circ, let us find \angle C:

\angle A+\angle B+\angle C = 180^\circ\\\Rightarrow 33+65.44+\angle C = 180\\\Rightarrow \angle C = 180-98.44 = 81.56^\circ

Let us find side c using sine rule again:

\dfrac{6}{sin33}=\dfrac{c}{sin81.56^\circ}\\\Rightarrow c  = 11.02 \times sin81.56^\circ = 10.89

So, one possible triangle is:

a = 6, b = 10, c = 10.89

\angleA=33°, \angleA=65.44°, \angleC=81.56°

For the value of \angle B =114.56^\circ, let us find \angle C:

\angle A+\angle B+\angle C = 180^\circ\\\Rightarrow 33+114.56+\angle C = 180\\\Rightarrow \angle C = 180-147.56 = 32.44^\circ

Let us find side c using sine rule again:

\dfrac{6}{sin33}=\dfrac{c}{sin32.44^\circ}\\\Rightarrow c  = 11.02 \times sin32.44^\circ = 5.91

So, second possible triangle is:

a = 6, b = 10, c = 5.91

\angleA=33°, \angleA=114.56°, \angleC=32.44°

So, answer is : 2 triangles are possible.

8 0
3 years ago
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