D) The measures of the non- right angles in each triangle .
Answer:

Step-by-step explanation:
The formula for the length of a vector/line in your case.
![L = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{[4 - (-1)]^2 + [2 -(-3)]^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}](https://tex.z-dn.net/?f=L%20%3D%20%5Csqrt%7B%28x_2-x_1%29%5E2%20%2B%20%28y_2-y_1%29%5E2%7D%20%3D%20%5Csqrt%7B%5B4%20-%20%28-1%29%5D%5E2%20%2B%20%5B2%20-%28-3%29%5D%5E2%7D%20%3D%20%5Csqrt%7B5%5E2%20%2B%205%5E2%7D%20%3D%20%5Csqrt%7B50%7D%20%3D%205%5Csqrt%7B2%7D)
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