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ddd [48]
3 years ago
15

How do you find p(a and b) ?

Mathematics
1 answer:
JulijaS [17]3 years ago
7 0

Answer:

Formula for the probability of A and B (independent events): p(A and B) = p(A) * p(B). If the probability of one event doesn't affect the other, you have an independent event. All you do is multiply the probability of one by the probability of another.

Step-by-step explanation:

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Find the system of x.​
marissa [1.9K]

Answer:

61 degrees

Step-by-step explanation:

A right angle equals 90 degrees.

90-29= 61

6 0
3 years ago
Read 2 more answers
You received a $178 per month raise. That was a 5% increase. What was your monthly salary before the increase? Show work please
zzz [600]

Answer: 169.52

Step-by-step explanation

%increase which is 5% =100 x final - initial

                                                   ---------------------

                                                         initial

the initial is $169.52

and the final  $178

3 0
3 years ago
What is the value of y?
Nataliya [291]

Answer:

B. 85

Step-by-step explanation:

I haven't taken Geometry in 2 years So I don't remember how to explain it.

5 0
3 years ago
PLZZ HELPP FIRST ANSWER THATS CORRECT GETS BRAINLIEST
Aleksandr-060686 [28]
Alright, so the sum of two days is going to equal zero.
The variable here is what Jesse got on the first day. Let's call it "x".
x-6=0
x is the first day's score, -6 is the second day's score, and the total score is zero.
Now, add 6 to both sides.
x-6=0
+6   +6
x=6
Jesse's score for the first day was (+)6.
5 0
3 years ago
Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
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