Simplify the expression

Hey!

To solve this problem, we would really have to graph both equations.

OPEN THE FIRST IMAGE

The first image I provided was the image of this equation graphed: y=2x+1

OPEN THE SECOND IMAGE

The second image I provided was the image of this equation graphed: y=2x^2+1

PLEASE DO NOT LOOK AT THE THIRD IMAGE YET!

The first image has two points. One at ( -0.5, 0 ) and the other at ( 0, 1 ).

The second image has only one point, which is at ( 0, 1 ).

OPEN THE THIRD IMAGE

Now, when both are combined we see that they intersect at two points, which are ( 0, 1 ) and ( 1, 3 ).

So, our answer is...

The ordered pairs that are the solution to the system are ( 0,1 ) and ( 1,3 ).

Hope this helped!

- Lindsey Frazier ♥

3 0

4 years ago

It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey

mote1985 [20]

Answer:

$P(X\geq 8)=0.0043\\\\$

It's more likely that all of the residents surveyed will have adequate earthquake supplies since it has a probability of 98.02% which is very close to 100%.

Step-by-step explanation:

-This is a binomial probability problem with the function:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}$

-Given p=0.3, n=11, the is calculated as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)+P(X=11)\\\\={11\choose 8}0.3^8(0.7)^3+{11\choose 9}0.3^9(0.7)^2+{11\choose 10}0.3^{10}(0.7)^1+{11\choose 11}0.3^{11}(0.7)^0\\\\=0.0037+0.0005+0.00005+0.000002\\\\=0.0043$

Hence, the probability that at least 8 have adequate supplies 0.0043

#The probability that non has adequate supplies is calculated as;

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X= 0)={11\choose 0}0.3^{0}(0.7)^{11}\\\\=0.0198$

#The probability that all have adequate supplies is calculated as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X= All)=1-{11\choose 0}0.3^{0}(0.7)^{11}\\\\=1-0.0198\\\\=0.9802$

Hence, it's more likely that all of the residents surveyed will have adequate earthquake supplies since $P(All)>P(None)\ \$ and that this probability is 0.9802 or 98.02% a figure close to 1

7 0

3 years ago