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kondor19780726 [428]
3 years ago
8

06.02)A six-sided number cube labeled 1 through 6 is rolled 600 times. An odd number is rolled 272 times. Compare the experiment

al probability of rolling an odd number with the relative frequency of rolling an odd number and select one of the statements below that best describes the situation.
The experimental probability and relative frequency are the same.
The experimental probability is larger than the relative frequency.
The experiment probability is smaller than the relative frequency.
There is not enough information to determine the relative frequency.
Mathematics
2 answers:
Elis [28]3 years ago
6 0

Answer:

Relative frequency is defined as the number of "desired" results ... A six-sided number cube labeled 1 through 6 is rolled 600 times. An odd number is rolled 272 times. Compare the experimental probability of rolling an odd number with ... of rolling an odd number

Step-by-step explanation:

horrorfan [7]3 years ago
4 0

Answer:

The experimental probability and relative frequency are the same.

Step-by-step explanation:

hope this helps!!!!!!!!!!

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A team of biologists captured and tagged 65 deer in a forest. Two weeks later, the
Gre4nikov [31]

Answer:

1/6x= 65

Step-by-step explanation:

so we can assume that 5/10= 1/6 of the deer were tagged

lets call the total number of dear x

so 1/6x= 65

x= 380

4 0
3 years ago
9th grade math please help
Leni [432]

Answer:

Only:

B. x=7

D. x=0

Step-by-step explanation:

Substitute the answers in the formula and only 7 and 0 will work.

6 0
3 years ago
Which of the following is true about the relation shown below?
Anestetic [448]

Answer:

Last on is the answer!! (The relation is a function, and the range is (-2,-1, 1, 3, 4). Hope this helps:)

6 0
2 years ago
Part A
masya89 [10]

Answer:

Part A) The area of triangle i is 3\ cm^{2}

Part B) The total area of triangles i and ii is 6\ cm^{2}

Part C) The area of rectangle i is 20\ cm^{2}

Part D) The area of rectangle ii is 32\ cm^{2}

Part E) The total area of rectangles i and iii is 40\ cm^{2}

Part F) The total area of all the rectangles is 72\ cm^{2}

Part G) To find the surface area of the prism, we need to know only the area of triangle i and the area of rectangle i and the area of rectangle ii, because the area of triangle ii is equal to the area of triangle i and the area of rectangle iii is equal to the area of rectangle i

Part H) The surface area of the prism is 78\ cm^{2}

Part I) The statement is false

Part J) The statement is true

Step-by-step explanation:

Part A) What is the area of triangle i?

we know that

The area of a triangle is equal to

A=\frac{1}{2} (b)(h)

we have

b=4\ cm

h=1.5\ cm

substitute

A=\frac{1}{2} (4)(1.5)

Ai=3\ cm^{2}

Part B) Triangles i and ii are congruent (of the same size and shape). What is the total area of triangles i and ii?

we know that

If Triangles i and ii are congruent

then

Their areas are equal

so

Aii=Ai

The area of triangle ii is equal to

Aii=3\ cm^{2}

The total area of triangles i and ii is equal to

A=Ai+Aii

substitute the values

A=3+3=6\ cm^{2}

Part C) What is the area of rectangle i?

we know that

The area of a rectangle is equal to

A=(b)(h)

we have

b=2.5\ cm

h=8\ cm

substitute

Ai=(2.5)(8)

Ai=20\ cm^{2}

Part D) What is the area of rectangle ii?

we know that

The area of a rectangle is equal to

A=(b)(h)

we have

b=4\ cm

h=8\ cm

substitute

Aii=(4)(8)

Aii=32\ cm^{2}

Part E) Rectangles i and iii have the same size and shape. What is the total area of rectangles i and iii?

we know that

Rectangles i and iii are congruent (have the same size and shape)

If rectangles i and iii are congruent

then

Their areas are equal

so

Aiii=Ai

The area of rectangle iii is equal to

Aiii=20\ cm^{2}

The total area of rectangles i and iii is equal to

A=Ai+Aiii

substitute the values

A=20+20=40\ cm^{2}

Part F) What is the total area of all the rectangles?

we know that

The total area of all the rectangles is

At=Ai+Aii+Aiii

substitute the values

At=20+32+20=72\ cm^{2}

Part G) What areas do you need to know to find the surface area of the prism?

To find the surface area of the prism, we need to know only the area of triangle i and the area of rectangle i and the area of rectangle ii, because the area of triangle ii is equal to the area of triangle i and the area of rectangle iii is equal to the area of rectangle i

Part H) What is the surface area of the prism? Show your calculation

we know that

The surface area of the prism is equal to the area of all the faces of the prism

so

The surface area of the prism is two times the area of triangle i plus two times the area of rectangle i plus the area of rectangle ii

SA=2(3)+2(20)+32=78\ cm^{2}

Part I) Read this statement: “If you multiply the area of one rectangle in the figure by 3, you’ll get the total area of the rectangles.” Is this statement true or false? Why?

The statement is false

Because, the three rectangles are not congruent

The total area of the rectangles is 72\ cm^{2} and if you multiply the area of one rectangle by 3 you will get 20*3=60\ cm^{2}

72\ cm^{2}\neq 60\ cm^{2}

Part J) Read this statement: “If you multiply the area of one triangle in the figure by 2, you’ll get the total area of the triangles.” Is this statement true or false? Why?

The statement is true

Because, the triangles are congruent

8 0
3 years ago
Every time I use a piece of scrap paper, I crumple it up and try to shoot it inside the recycling bin across the room. I'm prett
valentinak56 [21]

Using the binomial distribution, it is found that there is a 0.0012 = 0.12% probability at least two of them make it inside the recycling bin.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

With 5 shoots, the probability of making at least one is \frac{211}{243}, hence the probability of making none, P(X = 0), is \frac{232}{243}, hence:

(1 - p)^5 = \frac{232}{243}

\sqrt[5]{(1 - p)^5} = \sqrt[5]{\frac{232}{243}}

1 - p = 0.9908

p = 0.0092

Then, with 6 shoots, the parameters are:

n = 6, p = 0.0092.

The probability that at least two of them make it inside the recycling bin is:

P(X \geq 2) = 1 - P(X < 2)

In which:

[P(X < 2) = P(X = 0) + P(X = 1)

Then:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{6,0}.(0.0092)^{0}.(0.9908)^{6} = 0.9461

P(X = 1) = C_{6,1}.(0.0092)^{1}.(0.9908)^{5} = 0.0527

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.9461 + 0.0527 = 0.9988

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9988 = 0.0012

0.0012 = 0.12% probability at least two of them make it inside the recycling bin.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

7 0
2 years ago
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