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Lapatulllka [165]
3 years ago
14

Which two events are most likely to be independent? A) Being a senior; going to homeroom. B) Registering to vote; being left-han

ded. C) Having a car accident; having a junior license. D) Doing the Statistics homework; getting an A on the test.
Mathematics
2 answers:
Contact [7]3 years ago
6 0

Answer:

B,C

Step-by-step explanation:

those are unique to certain people

tankabanditka [31]3 years ago
6 0

Answer:

It is B) Registering to vote; being left-handed.

Step-by-step explanation:

These two events would be the only situation where the occurrence of one has no effect on the other, and each one of the other examples contain an event that affects the other probability of the other event.

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There are 36 students in Mrs. Keaton's sixth grade class.   If 5/12 of her students are girls, how many girls are in the class?
Gnesinka [82]
Multiply\ fraction\ by\ number\ of\ students\\\\\ number\ of\ girls=\frac{5}{12}*36=\frac{5*36}{12}=15\\\\\ There\ are\ 15\ girls\ in\ class.
3 0
4 years ago
The number of surface flaws in a plastic roll used for auto interiors follows a Poisson distribution with a mean of 0.05 flaw pe
ElenaW [278]

Answer:

0.6065

Step-by-step explanation:

Probability mass function of probability distribution : P(X=x)=\frac{e^{-\lambda} \times \lambda^x}{x !}

a mean of 0.05 flaw per square foot

Each car contains 10 sq.feet of the plastic roll

Mean = 0.05

Mean = \lambda = 0.05 \times 10=0.5

We are supposed to find What is the probability that there are no flaws in a given car’s interior i.e,P(X=0)

Substitute the value in the formula

P(X=0)=\frac{e^{-0.5} \times (0.5)^0x}{0 !}

P(X=0)=\frac{e^{-0.5} \times (0.5)^0}{1}

P(X=0)=0.6065

Hence the probability that there are no flaws in a given car’s interior is 0.6065

5 0
3 years ago
The diagram was constructed with straightedge and compass tools. Point A is the center of one circle, and point C is the center
Murrr4er [49]

Answer:

CE = AD/2

Step-by-step explanation:

C is the center of small circle

  • Point E is on same circle, therefore CE = radius of small circle
  • AD passes through the center of small circle and both points are on same circle, therefore AD is diameter of small circle

<u>Since diameter is double of radius:</u>

  • AD = d, CE= r
  • d = 2r
  • AD = 2*CE or CE = AD/2
5 0
3 years ago
How would I simplify and solve for X in this trigonometric equation? (Radians)
Phoenix [80]

Answer:

Please, refer to the images below

Step-by-step explanation:

We need to solve for x in the equation

cos (x+ pi) ^2 = sin (x)

cos (x+ pi) = - cos(x)

(-cos (x)) * (-cos (x)) = sin(x)

cos(x) ^2 = sin(x)

We know that

cos(x) ^2  + sin(x) ^2  = 1

cos(x) ^2 = 1 - sin(x) ^2

1 - sin(x) ^2 = sin(x)

sin(x) ^2 + sin (x) -1 = 0

Let A = sin(x)

A^2 + A - 1 = 0

(solutions attached in picture 1)

This means that

x = arcsin(A)

(solutions attached in picture 2)

8 0
3 years ago
The table shows the outputs y for different inputs x:
kirill [66]
Part A:

For a table to be considered a function, every x-value must have one y-value. 
Each x-value in this table is unique, and has only one y-value, so this table does represent a function.

Part B:

Plug in 7 for every x in the relation:

2(7) + 13 = 14 + 13 = 27

The table's output when x = 7 is 11. Compare the two outputs:

11 < 27

The relation, 2x + 13, has a greater value when x = 7.

Part C:

Set the relation to equal 75:

2x + 13 = 75

Subtract 13 from both sides:

2x = 62

Divide both sides by 2 to get x by itself:

\boxed{x = 31}

The x value that produces an output of 75 will be 31.
4 0
4 years ago
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