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Alla [95]
3 years ago
12

In a group of 65 students when you have brown eyes and 30 have blue eyes. Find the Probability that a student pick from this gro

up at random either has blue or brown eyes
Mathematics
1 answer:
Gemiola [76]3 years ago
4 0

Answer:

The probability that a student pick from this group at random either has blue or brown eyes is

P = 40/65 = 0.6154

P = 61.54%

Corrected question;

In a group of 65 students 10 have brown eyes and 30 have blue eyes. Find the Probability that a student pick from this group at random either has blue or brown eyes

Step-by-step explanation:

Given;

Total number of students = 65

Number of students with either blue or brown eyes;

= 10+30 = 40

The probability that a student pick from this group at random either has blue or brown eyes is

P = 40/65 = 0.6154

P = 61.54%

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Question 9 (1 point)
Romashka [77]

Answer:

Step-by-step explanation:

League A                   League B

151.12                            163.25

148                                157

26.83                             24.93

29                                  136

136                                145

167                                178

207                               256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}==\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

(82.5-1.5\times 76.56,159.06+1.5\times 76.56)

(-32.34,273.9)

So, There is no outlier

Maximum = 207

2)

League B in ascending order :

24.93,136,145,157,163.25,178,256

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}==\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

(140.5-1.5\times 30.125,170.625+1.5\times 30.125)

(95.3125,215.8125)

24.93 and 256 are outliers  

Maximum = 256

5 0
3 years ago
Algebra 2!..!!!! please help
dem82 [27]
We need to multiply the factors (x - 2) and (x^2 + 9x + 10) to see if the product is the original expression given.

(x - 2)(x^2 + 9x + 10)

x^3 + 9x^2 + 10x - 2x^2 - 18x - 20

x^3 + 7x^2 - 8x - 20.

Since the product just found is not the original expression given, Jimmy is wrong.

I will complete the work on paper by synthetic division and post my answer.

After using synthetic division, the correct quotient is x^2 + 9x + 20.
5 0
3 years ago
HELP PlS ASAP
Anarel [89]

Answer:

The correct answer is option B.

Step-by-step explanation:

Two point form of the equation:

A line passing through the point (-1,6) with slope ,m = -3.

The equation of the line will be: y-6= (-3)(x-(1))=(-3)(+1)

3 0
2 years ago
The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives 20.8 years; the
Luda [366]
If the mean is 20.8, one standard deviation each way is adding and subtracting 3.1, so 17.7 and 23.9 (68% of values)

Two standard deviations adding and subtracting 3.1*2 = 6.2, or 14.6 and 27.

Three standard deviations is 11.5 and 30.1

So we have
11.5 - 14.6 - 17.7 - 20.8 - 23.9 - 27 - 30.1

Going left to 11.5 is 3 standard deviations out, so 99.7/2 = 49.85%

Going right to 27 is 2 standard deviations out, so 95/2 = 47.5%

Add those two % to get 97.32%

This is hard to do without a picture so I hope that helps!

3 0
3 years ago
For the school play, 40 rows of chairs are set up. There are 22 chairs on each row. How many chairs are there?
Nataliya [291]
There are 880 chairs.
40x22=880
4 0
3 years ago
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