Change 2/4 to 1/2
<span><span>2x+<span>1/2</span>=189.5</span></span>
Subtract <span>1/2</span> from both sides(1/2 = .5)
2x=189
Divide both sides by <span>2</span>
2x/2 = x
189/2= 94.5
x=94.5
Answer: x>-6
Explanation: -2x -6>-18
-2x >-12
-x>6
X>-6
Answer: The mean and standard deviation are 567.2 and 89.88 resp.
Step-by-step explanation:
Since we have given that
For 370 parts per million = 7% = 0.07
For 440 parts per million = 10% = 0.10
For 550 parts per million = 49% = 0.49
For 670 parts per million = 34% = 0.34
So, Mean of the carbon dioxide atmosphere for these trees would be
![E[x]=370\times 0.07+440\times 0.1+550\times 0.49+670\times 0.34=567.2](https://tex.z-dn.net/?f=E%5Bx%5D%3D370%5Ctimes%200.07%2B440%5Ctimes%200.1%2B550%5Ctimes%200.49%2B670%5Ctimes%200.34%3D567.2)
And
![E[x^2]=370^2\times 0.07+440^2\times 0.1+550^2\times 0.49+670^2\times 0.34=329794](https://tex.z-dn.net/?f=E%5Bx%5E2%5D%3D370%5E2%5Ctimes%200.07%2B440%5E2%5Ctimes%200.1%2B550%5E2%5Ctimes%200.49%2B670%5E2%5Ctimes%200.34%3D329794)
So, Variance would be
![Var\ x=E[x^2]-E[x]^2=329794-567.2^2=8078.16](https://tex.z-dn.net/?f=Var%5C%20x%3DE%5Bx%5E2%5D-E%5Bx%5D%5E2%3D329794-567.2%5E2%3D8078.16)
So, the standard deviation would be

Hence, the mean and standard deviation are 567.2 and 89.88 resp.
Answer:
Felines: 40%
Reptiles: 25%
Aquatic Animals is 35%
Which means the Answer is 35%
(How you do this)
- First, Divide, the Numerator, by the Denominator and multiply it by 100.

=

Multiply both sides by v
<span>

= w Multiply both sides by x
</span>kv = wx Divide both sides by w
<span>

= x Switch the sides to make it easier to read
</span>
x = <span>
</span>