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Daniel [21]
2 years ago
8

The faces on a number cube are labled 1,2,2,3,4, and 5the number cube is rolled 114 times how many times would you expect the nu

mber 2 to apper
Mathematics
1 answer:
Mrac [35]2 years ago
4 0

Given:

The faces on a number cube are labeled 1,2,2,3,4, and 5.

The number cube is rolled 114 times.

To find:

How many times would you expect the number 2 to appear?

Solution:

We have,

Total outcomes = 1,2,2,3,4, and 5.

Number of total outcomes = 6

Favorable outcomes = 2 and 2

Number of favorable outcomes = 2

The probability of getting 2 is:

P(2)=\dfrac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}}

P(2)=\dfrac{2}{6}

P(2)=\dfrac{1}{3}

Now, the expected number of times when 2 to appear is:

E(x)=114\times P(2)

E(x)=114\times \dfrac{1}{3}

E(x)=38

Therefore, the expected number of times is 38.

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Can you add an image of the points on a graph so I can help?

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3 years ago
Subtract (8x – 2) – (5х – 7)
professor190 [17]

Answer: 3x - 9

Step-by-step-explanation: To solve the expression (8x – 2) – (5х – 7) we will be using properties of operations. First, use the commutative and associative properties to move the values next to each other to combine like terms.

(8x – 2) – (5х – 7)

(8x-5x)+(-2-7)

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So, the subtracted answer is 3x - 9

7 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20-%20x%20%5E%7B3%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ - x ^{3} } " alt=
noname [10]

I'm guessing you're given the function y(x)=2-x^3, and you're asked to find the inverse function y^{-1}(x). To do this, swap x and y, then solve for y:

x=2-y^3\implies y^3=2-x\implies y=(2-x)^{1/3}=\sqrt[3]{2-x}

so that the inverse function is

y^{-1}(x)=\sqrt[3]{2-x}

Just to verify:

y(y^{-1}(x))=y(\sqrt[3]{2-x})=2-(\sqrt[3]{2-x})^3=2-(2-x)=x

y^{-1}(y(x))=y^{-1}(2-x^3)=\sqrt[3]{2-(2-x^3)}=\sqrt[3]{x^3}=x

But in case you're actually only interested in computing the square root, first we note that \sqrt x (the real-valued square root) is only defined as long as x\ge0. So \sqrt{-x^3} is defined as long as -x^3\ge0, or x^3\le0, or equivalently x\le0. Under this condition, we could write

\sqrt{-x^3}=\sqrt{-x\times x^2}=\sqrt{-x}\sqrt{x^2}

We can simplify this further, but we have to be careful. Suppose x=-1. Then x^2=(-1)^2=1. But we get the same result if x=1, since x^2=1^2=1. There are two possible values of x that given the same value of x^2, so to capture both of them, we take \sqrt{x^2}=|x|, the absolute value of x. Then

\sqrt{-x^3}=|x|\sqrt{-x}

We can't simplify the square root term further than this.

3 0
3 years ago
When a boy pulls his sled with a rope, the rope makes an angle of 70° with the horizontal. If a pull of 20 pounds on the rope is
vladimir1956 [14]

Answer:

The answer to your question is 7 lb

Step-by-step explanation:

Data

angle = 70°

force = 20 lb

horizontal component

Process

To solve this problem just imagine a right triangle, where the force is the hypotenuse and the horizontal component is the adjacent side.

1.- Look for a trigonometric function that relates the adjacent side and the hypotenuse.

                   cos Ф = adjacent side / hypotenuse

- Solve for the adjacent side

                   adjacent side = hypotenuse x cosФ

- Substitution

                   adjacent side = 20 x cos 70

- Simplification

                   adjacent side = 20 x 0.342

- Result

                   adjacent side = 6.84 ≈ 7 lb  

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NISA [10]
The answer is A confirming appointments for the director of the facility
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2 years ago
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