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bija089 [108]
3 years ago
15

How do I solve for x

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
7 0
Try multiplying all of your numbers together
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klio [65]

Step-by-step explanation:

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7 0
3 years ago
What is the value of b2 - 4ac for the following equation? 2x 2 - 2x - 1 = 0
Orlov [11]
For you to make an equation a quadratic equation, you need everything to be on one side, so you add one to both sides. 

<span>2x^2 + 3x + 0=0 </span>

<span>a=2, b=3, c=0 </span>

<span>b^2-4ac (In the next step just plug in your variables) </span>

<span>3^2 - 4(2)(0) </span>

<span>9 - 8(0) </span>

<span>9 - 0 </span>

<span>Your answer=9</span>
6 0
3 years ago
Help please on this question!!
Alika [10]

Answer:

82

Step-by-step explanation:

50+32 =82 and this has to be longer so this is me making it longer

4 0
3 years ago
Read 2 more answers
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
Rewrite the following logarithmic expression as an exponential expression
Alina [70]

Answer:

it should be the same answer....?

Step-by-step explanation:

3 0
3 years ago
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