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Makovka662 [10]
3 years ago
7

Why could the mean of the data set below be misleading

Mathematics
1 answer:
Alika [10]3 years ago
3 0
Because of the outlier 77
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Which of the following fractions is not equivalent to 8/22 ?
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B because 5xanything ends in 0 or 5
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3 years ago
Find the factors of 36
Gemiola [76]

Hello,


<em>Factors: 1, 2, 3, 4, 6, 9, 12, 18, 36 </em>

<em> </em>

<em>Factor Pairs: (1, 36) (2, 18) (3, 12) (4, 9) (6, 6) </em>

<em> </em>

<em>Prime factors: 36 = 2 × 2 × 3 × 3 </em>


Thanks,

<em>Deku ❤</em>

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3 years ago
wo balls are chosen randomly from an um containing 8 white, 4 black,and 2 orange balls. Suppose that we win $2 for each black ba
umka21 [38]

Answer:

The probability distribution is shown below.

Step-by-step explanation:

The urn consists of 8 white (<em>W</em>), 4 black (<em>B</em>) and 2 orange (<em>O</em>) balls.

The winning and losing criteria are:

  • Win $2 for each black ball selected.
  • Lose $1 for each white ball selected.

There are 8 + 4 + 2 = 14 balls in the urn.

The number of ways to select two balls is, {14\choose 2}=91 ways.

The distribution of amount won or lost is as follows:

Outcomes: WW  WO  WB  BB  BO  OO

X:                 -2      -1      1      4     2      0

Compute the probability of selecting 2 white balls as follows:

The number of ways to select 2 white balls is, {8\choose 2}=28 ways.

The probability of WW is,

P(WW)=\frac{n(WW)}{N}=\frac{28}{91}=0.3077

Compute the probability of selecting 1 white ball and 1 orange ball as follows:

The number of ways to select 1 white ball and 1 orange ball is, {8\choose 1}\times {2\choose 1}=16 ways.

The probability of WO is,

P(WO)=\frac{n(WO)}{N}=\frac{16}{91}=0.1758

Compute the probability of selecting 1 white ball and 1 black ball as follows:

The number of ways to select 1 white ball and 1 black ball is, {8\choose 1}\times {4\choose 1}=32 ways.

The probability of WB is,

P(WB)=\frac{n(WB)}{N}=\frac{32}{91}=0.3516

Compute the probability of selecting 2 black balls as follows:

The number of ways to select 2 black balls is, {4\choose 2}=6 ways.

The probability of BB is,

P(BB)=\frac{n(BB)}{N}=\frac{6}{91}=0.0659

Compute the probability of selecting 1 black ball and 1 orange ball as follows:

The number of ways to select 1 black ball and 1 orange ball is, {4\choose 1}\times {2\choose 1}=8 ways.

The probability of BO is,

P(BO)=\frac{n(BO)}{N}=\frac{8}{91}=0.0879

Compute the probability of selecting 2 orange balls as follows:

The number of ways to select 2 orange balls is, {2\choose 2}=1 ways.

The probability of OO is,

P(OO)=\frac{n(OO)}{N}=\frac{1}{91}=0.0110

The probability distribution of <em>X</em> is:

Outcomes:    WW     WO        WB         BB        BO         OO

X:                    -2          -1            1            4            2            0

P (X):           0.3077  0.1758  0.3516  0.0659  0.0879  0.0110

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4 years ago
Karen bought a total of seven items at five different stores. She began with $65.00 and had $15.00 remaining. Which of the follo
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7 0
4 years ago
Jacob won 62 super bouncy balls playing
Mkey [24]

Answer:

30

Step-by-step explanation:

62 = 2x + 2

62 - 2 = 2x

60 = 2x

x = 30

8 0
3 years ago
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