You can write it in the form a^2 - b^2 where a = 5x and b = p
(5x)^2 - p^2
And then use the Difference of Squares a^2 - b^2 = (a + b)(a - b)
Answer is (5x + p)(5x - p)
Answer:
2 x <u>2</u> + 9 > 11
2 x 2 = 4
4 + 9 = 13
13 > 11
Therefore, x = 2.
Step-by-step explanation:
I hope this helps! ^w^
<span>4.a translation 4 units right and 4 units down</span>
Answer:
just break it down and you will get it! :D
Step-by-step explanation:
Answer:
C. 5 weeks.
Step-by-step explanation:
In this question we have a random variable that is equal to the sum of two normal-distributed random variables.
If we have two random variables X and Y, both normally distributed, the sum will have this properties:
![S=X+Y\\\\\ \mu_S=\mu_X+\mu_Y=30+60=90\\\\\sigma_S=\sqrt{\sigma_X^2+\sigma_Y^2}=\sqrt{10^2+20^2}=\sqrt{100+400}=\sqrt{500}=22.36](https://tex.z-dn.net/?f=S%3DX%2BY%5C%5C%5C%5C%5C%20%5Cmu_S%3D%5Cmu_X%2B%5Cmu_Y%3D30%2B60%3D90%5C%5C%5C%5C%5Csigma_S%3D%5Csqrt%7B%5Csigma_X%5E2%2B%5Csigma_Y%5E2%7D%3D%5Csqrt%7B10%5E2%2B20%5E2%7D%3D%5Csqrt%7B100%2B400%7D%3D%5Csqrt%7B500%7D%3D22.36)
To calculate the expected weeks that the donation exceeds $120, first we can calculate the probability of S>120:
![z=\frac{S-\mu_S}{\sigma_S} =\frac{120-90}{22.36}=\frac{30}{22.36}=1.34\\\\P(S>120)=P(z>1.34)=0.09012](https://tex.z-dn.net/?f=z%3D%5Cfrac%7BS-%5Cmu_S%7D%7B%5Csigma_S%7D%20%3D%5Cfrac%7B120-90%7D%7B22.36%7D%3D%5Cfrac%7B30%7D%7B22.36%7D%3D1.34%5C%5C%5C%5CP%28S%3E120%29%3DP%28z%3E1.34%29%3D0.09012)
The expected weeks can be calculated as the product of the number of weeks in the year (52) and this probability:
![E=\#weeks*P(S>120)=52*0.09012=4.68](https://tex.z-dn.net/?f=E%3D%5C%23weeks%2AP%28S%3E120%29%3D52%2A0.09012%3D4.68)
The nearest answer is C. 5 weeks.