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2 years ago

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Two events, X and Y, are independent of each other. P(Y) = 5/6 and P(X and Y) = 1/3. What is P(X) written as a decimal? Round to

the nearest tenth, if necessary.

2 answers:

Wittaler [7]2 years ago

7 0

Answer:

answer is 0.4

Step-by-step explanation:

Ksenya-84 [330]2 years ago

5 0

For independent events,
P(X ∩ Y) = P(X)* P(Y)
=>
1/3 = P(X)*(5/6)
solve for P(X) =>
P(X) = (1/3)*(6/5) = 2/5 = 0.4

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Solve this using substitution:<br> 5x-y=-3<br> 6x+y=3

Marta_Voda [28]

Answer:

x=0.y=3

Step-by-step explanation:

5x-y=-3

y=5x+3

sub y=5x+3 into 6x+y=3,

6x+5x+3=3

11x=0

x=0

6(0)+y=3

y=3

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2 years ago

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Determine the factors of 3x2 + 10x − 8. (Which is the right answer?)

VMariaS [17]

ANSWER

$(3x - 2)(x + 4)$

EXPLANATION

The given quadratic trinomial is

$3 {x}^{2} + 10x - 8$

Comparing to the general quadratic trinomial:

$a{x}^{2} + bx + c$

We have a=1, b=10 and c=-8

$ac = 3 \times - 8 = - 24$

The factors of -24 that adds up to 10 are -2 and 12.

We split the middle term to obtain;

$3 {x}^{2} + 12x - 2x- 8$

Factor;

$3x(x+ 4) - 2(x + 4)$

Factor further;

$(3x - 2)(x + 4)$

The last option is correct.

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3 years ago

If you flip three fair coins what is the probability that the two first flips will both be heads, and the third flip will be eit

Brrunno [24]

Answer:

The probability that a coin flip is a head or a tail is 0.5. The probability that a coin flip is either a head or a tail is 1. The probability that a three-coin-flip is two heads and a head or a tail is 0.25.

Step-by-step explanation:

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2 years ago

A student’s height is 145cm. He goes to a pond where the average water depth is 110cm. Does this student have a risk in the pond

Elodia [21]

Yes, because 145-110=35

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2 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

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2 years ago