Answer: 0.0475
Step-by-step explanation:
Let x = random variable that represents the number of a particular type of bacteria in samples of 1 milliliter (ml) of drinking water, such that X is normally distributed.
Given: 
The probability that a given 1-ml will contain more than 100 bacteria will be:
![P(X>100)=P(\dfrac{X-\mu}{\sigma}>\dfrac{100-85}{9})\\\\=P(Z>1.67)\ \ \ \ [Z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Zz)=1-P(Z](https://tex.z-dn.net/?f=P%28X%3E100%29%3DP%28%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cdfrac%7B100-85%7D%7B9%7D%29%5C%5C%5C%5C%3DP%28Z%3E1.67%29%5C%20%5C%20%5C%20%5C%20%5BZ%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-P%28Z%3C1.67%29%5C%20%5C%20%5C%20%5BP%28Z%3Ez%29%3D1-P%28Z%3Cz%29%5D%5C%5C%5C%5C%3D1-%200.9525%3D0.0475)
∴The probability that a given 1-ml will contain more than 100 bacteria
0.0475.
Solve the system of equations.
12x + 3y = 12
-6x +y = -26
A) (-3,9)
B (-6, -9)
© (-3,8)
D (3, -8)
12x + 3y = 12
2(-6x +y = -26)
12x + 3y = 12
-12x + 2y = -52
—-now x cancels out
5y/5 = -40/5
y = -8
The answer should be D. Let me know if you got it right.
Answer:
7⅓
Step-by-step explanation:
2 ÷ 3⁄11 → 2 × 3⅔ >> 7⅓
Dividing by a fraction is the exact same as multiplying by its multiplicative inverse.
* 3⅔ = 11⁄3
* 7⅓ = 22⁄3
I am joyous to assist you anytime.
Answer:
$49.60
Step-by-step explanation:
The equation setup I used for this was as follows: $16 x (112/36)
Using PEMDAS, you'd start with the division part, 112/36. Although, I did this a bit differently, and being a bit further in school, I don't know if you go through this where you're at.
Start by factoring 112 into 4(28). This leaves you with 4(28)/36.
Next, cancel 4 out of 36. Since 4 is the common factor in 28 and 36, you cancel the four out of the equation, leaving you with 28/9. Convert this to its decimal form, 3.1.
Finally, take the 3.1 and multiply that by $16, which comes out to $49.60.
