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3 years ago

Find the distance between the points (–1∕2,–8) and (–1∕2,–2). Question 5 options: A) 6 B) 10 C) 8 D) 1∕2

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

3 0

Answer:

A) 6

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra II

Distance Formula: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Step-by-step explanation:

Step 1: Define

Point (-1/2, -8)

Point (-1/2, -2)

Step 2: Find distance d

Substitute: $d = \sqrt{(\frac{-1}{2} +\frac{1}{2} )^2+(-2+8)^2}$
Add: $d = \sqrt{(0)^2+(6)^2}$
Exponents: $d = \sqrt{0+36}$
Add: $d = \sqrt{36}$
Evaluate: $d = 6$

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Two consecutive positive integers have a product of 6 what are the integers

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Answer:

2,3

Step-by-step explanation:

2 and 3 are consecutive numbers, and when you multiply them, you get 6.

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3 years ago

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Alona [7]

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3 years ago

A player tosses a die 6 times.If gets a number 6 Atleast two times he wins 2 dollars ,otherwise he looses 1 dollar.. Find the ex

swat32

Answer:

$E(x)=-0.2101$

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

$E(x)=x_1*p(x_1)+x_2*p(x_2)$

Where $x_1$ and $x_2$ are the values that the variable can take and $p(x_1)$ and $p(x_2)$ are their respective probabilities.

So, a player can win 2 dollars or looses 1 dollar, it means that $x_1$ is equal to 2 and $x_2$ is equal to -1.

Then, we need to calculated the probability that the player win 2 dollars and the probability that the player loses 1 dollar.

If there are n identical and independent events with a probability p of success and a probability (1-p) of fail, the probability that a events from the n are success are equal to:

$P(a)=nCa*p^a*(1-p)^{n-a}$

Where $nCa=\frac{n!}{a!(n-a)!}$

So, in this case, n is number of times that the player tosses a die and p is the probability to get a 6. n is equal to 6 and p is equal to 1/6.

Therefore, the probability $p(x_1)$ that a player get at least two times number 6, is calculated as:

$p(x_1)=P(x\geq2)=1-P(0)-P(1) \\\\P(0) =6C0*(1/6)^{0}*(5/6)^{6}=0.3349\\P(1)=6C1*(1/6)^{1}*(5/6)^{5}=0.4018\\\\p(x_1)=1-0.3349-0.4018\\p(x_1)=0.2633$

On the other hand, the probability $p(x_2)$ that a player don't get at least two times number 6, is calculated as:

$p(x_2)=1-p(x_1)=1-0.2633=0.7367$

Finally, the expected value of the amount that the player wins is:

$E(x)=x_1*p(x_1)+x_2*p(x_2)\\E(x)=2*(0.2633)+ (-1)*0.7367\\E(x)=-0.2101$

It means that he can expect to loses 0.2101 dollars.

3 0

3 years ago

Read 2 more answers

Fill in the blanks:-

aleksley [76]

Answer:

The product of a number and its reciprocal is 1.

For i.e : 4/3 and its reciprocal 3/4, 4/3 x 3/4 = 1

Hope this helps!

6 0

3 years ago