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Blababa [14]
2 years ago
15

Plz help!! A jar contained 6 red marbles, 10 green marbles, and 8 blue marbles. A marble was selected at random, the color was r

ecorded, and the marble was placed back in the jar. The table shows the results after 60 trials. What was the relative frequency of selecting a red marble? Outcome Number of times outcome occurred Red 13 Green 30 Blue 17
Give your answer as a fraction in simplest form.
Mathematics
1 answer:
Inessa [10]2 years ago
3 0

Answer:

To calculate the relative frequency, first we need to know what exactly is and how to calculate it.

Relative frequency is the ratio between the absolute frequency (how many repetitions have a specific outcome) and the total outcomes. Also, this type of frequency is used to show the representation that some data have over the whole distribution.

So, in this case, we need to just divide 13, which belongs to red marble's results, to 60 which is the total outcomes, as it's presented:

         13redmarble                                                                                                                       Fr = -------------------------

       60 totalmarbles  

Normally, relative frequency is shown as a percentage multiplying this result by 100. Therefore, 22% is the approximate percentage of the relative frequency, which means that 22% is the representation of red marbles outcomes of the whole distribution, or we can say it as a probability: there's 22% of chances when someone extract a marble, it will be red.

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PLEASE HELP W/ THIS QUESTION I ALLWAYS GET STUCK ON READING TIME!!!
Ad libitum [116K]

Answer:

2:40

Step-by-step explanation:

Short hand (hours): a little past 2

Long hand (minutes): 40 -- 8*5=40

6 0
2 years ago
A sports survey taken at THS shows that 48% of the respondents liked soccer, 66% liked basketball and 38% liked hockey. Also,30%
In-s [12.5K]

Answer:

a) The Venn diagram is presented in the attached image to this answer.

b) 0.82

c) 0.16

Step-by-step explanation:

a) The Venn diagram is presented in the attached image to this answer.

n(U) = 100%

n(S) = 48%

n(B) = 66%

n(H) = 38%

n(S n B) = 30%

n(B n H) = 22%

n(S n H) = 28%

n(S n B n H) = 12%

The specific breakdowns for each subgroup is calculated on the Venn diagram attached.

b) The probability that a randomly selected student likes basketball or hockey.

P(B U H)

From the Venn diagram,

n(B U H) = n(S' n B n H') + n(S' n B n H) + n(S n B n H') + n(S n B n H) + n(S n B' n H) + n(S' n B' n H) = 26 + 10 + 18 + 12 + 16 + 0 = 82%

P(B U H) = 82/100 = 0.82

c) The probability that a randomly selected student does not like any of these sports.

P(S' n B' n H')

n(S' n B' n H') = n(U) - [n(S' n B n H') + n(S' n B n H) + n(S n B n H') + n(S n B n H) + n(S n B' n H) + n(S' n B' n H) + n(S n B' n H')]

n(S' n B' n H') = 100 - (26 + 10 + 18 + 12 + 16 + 0 + 2) = 100 - 84 = 16%

P(S' n B' n H') = 16/100 = 0.16

4 0
3 years ago
Can somebody please walked through this, I'm so confused and I have a test in 6 hours...
erastova [34]

Answer:

Q3: x = 4, y = 4, z = 4

Q4: x = 6, y = 0, z = -4

Step-by-step explanation:

Question 3: Simultaneous equations requires us to solve for x, y and z.

Since all three equations have a z in them, I will first solve for z.

Substitute in the first and third equation into the second equation.

First equation: x = 5z - 16

Second Equation: -4x + 4y - 5z = -20

Third equation: y = -z + 8

Substituting in x = 5z - 16 and y = -z + 8 for the x and y in the second equation.

-4(5z - 16) + 4(-z + 8) - 5z = -20

Expand

-20z + 64   - 4z + 32   - 5z = -20

Simplify and solve for z by putting all the numbers on one side and all the z's on the other side of the equals

-20z - 4z - 5z = -20 - 32 - 64

-29z = -116

z = -116/-29

z = 4

Substitute in this z value into the first and last equation and then solve for x and y

x = 5z - 16

x = 5(4) - 16

x = 20 - 16

x = 4

And

y = -z + 8

y = -(4) + 8

y = 4 (Its just a coincidence that they all equal to 4, I promise)

Question 5: A little bit harder of a question. Since the first and second equation both only have y and z, we can solve it using the elimination method.

Rearrange them so that the letters are on one side and numbers on the other side.

First equation: y + 6z = -24

Second equation: z + 2y = -4

I will choose to eliminate the y (You can choose either or)

Multiply the first equation by 2

2(y + 6z = -24)

2y + 12z = -48

Now that 2y is in both equations, we can minus one equation from the other to eliminate the y (I will minus the second from the first)

First Eq: 2y + 12z = -48

Second Eq: z + 2y = -4

2y - 2y = 0y

12z - z = 11z

-48 - (-4) = -44

Type these answers into a new equation

0y + 11z = - 44

Since y is 0, ignore it. Solve for z

11z = -44

z = -44/11

z = - 4

Substitute our z into either the first or second equation and solve for y (It doesnt matter which one you choose, I just did the second equation)

z + 2y = -4

(-4) + 2y = -4

2y = -4 + 4

2y = 0

y = 0

Substitute in our y and z values into the third equation and solve for x

-6x - 6y - 6z = -12

-6x - 6(0) - 6(-4) = -12

-6x - 0 + 24 = -12

-6x = -12 - 24

-6x = -36

x = -36/-6

x = 6

6 0
3 years ago
Read 2 more answers
Find the indicated measure in parallelogram ABCD m
Zarrin [17]

Check the picture below.

those angles in red are alternate interior with the angles at A and D, therefore, equal.

8 0
3 years ago
Could someone help me out please
aev [14]

Answer:

Six Tenths in numerical form is 6/10, and "translated" into a decimal it would be 0.6

I hope I helped you, and a "Brainliest" is always appreciated! ☺

8 0
3 years ago
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