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3 years ago

How do u solve d÷12=2

Mathematics

1 answer:

garri49 [273]3 years ago

5 0

Just do 12 times 2 d=24 because 24 divided by 12 =2

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How can you convert units by capacity?

seraphim [82]

Answer:

division by whatever each is worth, hope this makes sense!

Step-by-step explanation:

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3 years ago

In a survey of sixth-grade students, 70 students, or 40% of all the students surveyed, said that they liked mustard on hot dogs.

n200080 [17]

Answer: 210

<em>Step-by-step explanation: 70+70+70=210 and 40% for 210 students would add up to 120% of all students surveyed. Look if there is 70 students that's been surveyed that is 40% add 40% 3 times(40%+40%+40%)=120% of all students surveyed </em>

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3 years ago

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Which of the following equations are equivalent to 2x(x - 2) = x? Select two that apply.

Verdich [7]

Answer: C 2x(x - 2) - x = 0

and <u>B</u> 2x² – 5x=0

Step-by-step explanation:

You get 2x(x - 2) - x = 0 by subtracting x from both sides of the original equation.

Then take 2x(x - 2) - x = 0 and distribute to get 2x² - 4x -x = 0

combine the x terma and you end up with 2x² – 5x=0

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2 years ago

A soccer team has 18 players. 5 of the players have scored at least 4 goals this season. Approximately what percent of the playe

liberstina [14]

Answer:30%

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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.

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3 years ago

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