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Verdich [7]
3 years ago
15

Let the sample space beUpper S equals StartSet 1 comma 2 comma 3 comma 4 comma 5 comma 6 comma 7 comma 8 comma 9 comma 10 EndSet

S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.Suppose the outcomes are equally likely. Compute the probability of the eventUpper E equals StartSet 1 comma 2 comma 4 comma 6 EndSetE={1, 2, 4, 6}.
Mathematics
1 answer:
AleksAgata [21]3 years ago
7 0

Answer:

The probability would be 0.40

Step-by-step explanation:

Given,

The elements in set S are,

1, 2, 3, 4, 5, 6, 7, 9, 10,

Number of elements = 10,

Since, the outcomes are equally likely,

And, we know that,

\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}

So, the probability of getting 1, 2, 4, or 6 = \frac{4}{10}=0.40

Hence, the probability of the event E = {1, 2, 4, 6} is 0.40.

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Callie brushes her teeth three times each day. She leaves the water running. Is it reasonable to say that she uses 2 cups of wat
Dmitriy789 [7]

Answer: She uses more than 2 cups of water per day.

Step-by-step explanation:

It takes 2 minutes to brush your teeth and if you times that by 3 you get 6 so she spends 6 minuntes per day brushing her teeth and water runs fast. so if your leaving the water on for 6 mintues your filling up much more than 2 cups

5 0
3 years ago
Thirty students at Eastside High School took the SAT on the same Saturday. Their raw scores are given next. 1,450 1,620 1,800 1,
IRISSAK [1]

Answer:

Number of students who scored atleast 1700 but less Than 2100 = 17

Step-by-step explanation:

Given the data :

1,450 1,620 1,800 1,740 1,650 1,800 2,010 1,780 1,840 1,490 1,620 1,480 2,390 1,640 1,830 1,710 1,900 1,910 1,950 1,820 1,590 2,350 2,260 1,870 1,530 1,950 2,000 1,830 1,980 2,100

Class interval __ Frequency

1400 - 1599 ____ 5

1600 - 1799 ____ 7

1800 - 1999 ____12

2000 - 2199 ___ 3

2200 - 2399 ___ 3

Number of students who scored atleast 1700 but less Than 2100 = 17

8 0
3 years ago
The following is a linear programming formulation of a labor planning problem. There are four overlapping shifts, and management
dexar [7]

Answer:

d. 15

Step-by-step explanation:

Putting the values in the shift 2 function

X1 + X2 ≥ 15

where x1=  13, and x2=2

13+12≥ 15

15≥ 15

At least 15 workers must be assigned to the shift 2.

The LP model questions require that the constraints are satisfied.

The constraint for the shift 2 is that the  number of workers must be equal or greater than 15

This can be solved using other constraint functions e.g

Putting  X4= 0 in

X1 + X4 ≥ 12

gives

X1 ≥ 12

Now Putting the value X1 ≥ 12  in shift 2 constraint

X1 + X2 ≥ 15

12+ 2≥ 15

14 ≥ 15

this does not satisfy the condition so this is wrong.

Now from

X2 + X3 ≥ 16

Putting X3= 14

X2 + 14 ≥ 16

gives

X2  ≥ 2

Putting these in the shift 2

X1 + X2 ≥ 15

13+2 ≥ 15

15 ≥ 15

Which gives the same result as above.

6 0
3 years ago
Given that F(x) = x2 + 2, evaluate F(1) + F(5).
gavmur [86]

Answer:

F(1)+F(5)= 30

Step-by-step explanation:

F(1)=1^2+2=1+2=3

F(5)=5^2+2=25+2=27

27+3=30

6 0
3 years ago
Geometry hellpppppo<br> 20 POINTS PLS HELP
sesenic [268]
Sin(x) = base / hypotenuse

sin(x) = 7/9

x = arcsin(7/9)

x = 51 degrees
5 0
3 years ago
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