Answer:
3496
Step-by-step explanation:
multiple for mine idea
Answer:
a) 0.50575,
b) 0.042
Step-by-step explanation:
Example 1.5. A person goes shopping 3 times. The probability of buying a good product for the first time is 0.7.
If the first time you can buy good products, the next time you can buy good products is 0.85; (I interpret this as, if you buy a good product, then the next time you buy a good product is 0.85).
And if the last time I bought a bad product, the next time I bought a good one is 0.6. Calculate the probability that:
a) All three times the person bought good goods.
P(Good on 1st shopping event AND Good on 2nd shopping event AND Good on 3rd shopping event) =
P(Good on 1st shopping event) *P(Good on 2nd shopping event | Good on 1st shopping event) * P(Good on 3rd shopping event | 1st and 2nd shopping events yield Good) =
(0.7)(0.85)(0.85) =
0.50575
b) Only the second time that person buys a bad product.
P(Good on 1st shopping event AND Bad on 2nd shopping event AND Good on 3rd shopping event) =
P(Good on 1st shopping event) *P(Bad on 2nd shopping event | Good on 1st shopping event) * P(Good on 3rd shopping event | 1st is Good and 2nd is Bad shopping events) =
(0.7)(1-0.85)(1-0.6) =
(0.7)(0.15)(0.4) =
0.042
Answer:
Step-by-step explanation:
(-9-7)/(0-1)= -16/-1= 16
y - 7 = 16(x - 1)
y - 7 = 16x - 16
y = 16x - 9
To find h(6), we must plug in 6 into the function.
h(x) = x^2 - 2x - 5
h(6) = 6^2 - 2(6) - 5
= 36 - 12 - 5
= 36 - 17 = 19
