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3 years ago

Plz help.....................

Mathematics

1 answer:

sdas [7]3 years ago

8 0

Perfect square trinomial means that the constant term is a perfect square, and the middle term is the sum of it's factors.
Option 1: constant 64 sqrt = 8; 8 + 8 = 16 or -8 + -8 = -16 NO
Option 2: constant 36 sqrt = 6; 6 + 6 = 12 or -6 + -6 = -12 NO
Option 3: constant 64 sqrt = 8; 8 + 8 = 16 or -8 + -8 = -16 YES
(x - 8)^2 is the factored form

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What is the measure of x?

bogdanovich [222]

Answer:

Step-by-step explanation:

This is a right angle so the sum of those would be equal to 90 degrees

x + 7 + 3x - 5 = 90 add like terms

4x + 2 = 90 subtract 2 from both sides

4x = 88 divide both sides by 4

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Determine whether this system of equations has one solution, no solution, or infinitely many solutions.

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The only solution I see is for x=0

Step-by-step explanation:

y=8x+2

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3 years ago

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The Third one

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Because the y is not included in the function ( see some lines ) and x is the same thing but it is larger than -3 and it is included in the function .

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3 years ago

How many inches are in three fourths of a yard

guapka [62]

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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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