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3 years ago

PLZ HELP fill in the blanks

Mathematics

2 answers:

Fantom [35]3 years ago

8 0

5 x (2 + 9) = (5 x 2) + (5 x 9)
Ok done. Thank to me :>

Alexxx [7]3 years ago

6 0

Answer:

Step-by-step explanation:

First, to solve the problem, you must distribute the 5 to the (2 + 9). Which ends you up with 10 + 45. Which would give you 55, but we aren't solving it.

So instead, when you are distributing, you are taking the number next to the parentheses (5) and multiplying it into the parentheses like this:

(5 x 2) + (5 + 9).

That is your answer. Hope this helps.

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A ___ is the result of a division problem.

natita [175]

Answer: The dividend is divided by the divisor to get the result which is the quotient.

3 0

3 years ago

Y+2=1/3 (x+1) please can you help

miskamm [114]

Equation:

y + 2 = 1/3(x+1)

Assuming you are solving for Y:

y + 2 = 1/3(x) + 1/3(1)

(y + 2) - 2 = (1/3x + 1/3) - 2

y = 1/3x + 1/3 - 2

7 0

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

What is the sum of the geometric series?

Dvinal [7]

Essentially this is asking you to sum the first three values of the geometric series that starts with i being 1 and has the rule of ( $8(\frac{1}{4})^{i-1}$ ).

Since we must sum the first <em>three </em>geometric values we need to see what the value of i will be for the first three. The sigma notation shows us that the first i will be equal to 1. This means the second i is 2 and the third i is 3.

Knowing this you can plug in the corresponding i values into ( $8(\frac{1}{4})^{i-1}$ ) and sum it all together

( $8(\frac{1}{4})^{1-1}$ ) + ( $8(\frac{1}{4})^{2-1}$ ) + ( $8(\frac{1}{4})^{3-1}$ )