M - 536 <== ur expression
when the mass is 781 grams...
781 - 536 = 245
so the mass of the package is 781....and when it is reduced by 536 grams, it equals 245 grams
Answer:
no se
Step-by-step explanation:
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Answer:
a = - 2, b = - 6
Step-by-step explanation:
Substitute the values of the zeros into the polynomial and equate to zero.
x² +(a + 1)x + b
x = - 2 → (- 2)² - 2(a + 1) + b = 0 , that is
4 - 2a - 2 + b = 0
2 - 2a + b = 0 ( subtract 2 from both sides )
- 2a + b = - 2 → (1)
x = 3 → 3² + 3(a + 1) + b = 0, that is
9 + 3a + 3 + b = 0
12 + 3a + b = 0 ( subtract 12 from both sides )
3a + b = - 12 → (2)
Subtract (1) from (2) term by term to eliminate b
5a = - 10 ( divide both sides by 5 )
a = - 2
Substitute a = - 2 into either of the 2 equations and evaluate for b
Substituting into (2)
3(- 2) + b = - 12
- 6 + b = - 12 ( add 6 to both sides )
b = - 6
Thus a = - 2 and b = - 6
Answer:
114 cm2
Step-by-step explanation:
Answer:
a) P(x<5)=0.
b) E(X)=15.
c) P(8<x<13)=0.3.
d) P=0.216.
e) P=1.
Step-by-step explanation:
We have the function:

a) We calculate the probability that you need less than 5 minutes to get up:

Therefore, the probability is P(x<5)=0.
b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.
E(X)=15.
c) We calculate the probability that you will need between 8 and 13 minutes:
![P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3](https://tex.z-dn.net/?f=P%288%5Cleq%20x%5Cleq%2013%29%3DP%2810%5Cleqx%5Cleq%2013%29%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cint_%7B10%7D%5E%7B13%7D%20f%28x%29%5C%2C%20dx%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cint_%7B10%7D%5E%7B13%7D%20%5Cfrac%7B1%7D%7B10%7D%20%5C%2C%20dx%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Ccdot%20%5Bx%5D_%7B10%7D%5E%7B13%7D%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Ccdot%20%2813-10%29%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cfrac%7B3%7D%7B10%7D%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D0.3)
Therefore, the probability is P(8<x<13)=0.3.
d) We calculate the probability that you will be late to each of the 9:30am classes next week:
![P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6](https://tex.z-dn.net/?f=P%28x%3E14%29%3D%5Cint_%7B14%7D%5E%7B20%7D%20f%28x%29%5C%2C%20dx%5C%5C%5C%5CP%28x%3E14%29%3D%5Cint_%7B14%7D%5E%7B20%7D%20%5Cfrac%7B1%7D%7B10%7D%20%5C%2C%20dx%5C%5C%5C%5CP%28x%3E14%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Bx%5D_%7B14%7D%5E%7B20%7D%5C%5C%5C%5CP%28x%3E14%29%3D%5Cfrac%7B6%7D%7B10%7D%5C%5C%5C%5CP%28x%3E14%29%3D0.6)
You have 9:30am classes three times a week. So, we get:

Therefore, the probability is P=0.216.
e) We calculate the probability that you are late to at least one 9am class next week:
![P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1](https://tex.z-dn.net/?f=P%28x%3E9.5%29%3D%5Cint_%7B10%7D%5E%7B20%7D%20f%28x%29%5C%2C%20dx%5C%5C%5C%5CP%28x%3E9.5%29%3D%5Cint_%7B10%7D%5E%7B20%7D%20%5Cfrac%7B1%7D%7B10%7D%20%5C%2C%20dx%5C%5C%5C%5CP%28x%3E9.5%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Bx%5D_%7B10%7D%5E%7B20%7D%5C%5C%5C%5CP%28x%3E9.5%29%3D1)
Therefore, the probability is P=1.