Answer:
=14
Step-by-step explanation:
...................
The minimum Sam need to get is a 70.
70 + 60 = 130
130/2 = 65
Answer:
a. attached graph; zero real: 2
b. p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)
c. the solutions are 2, -3-3i and -3+3i
Step-by-step explanation:
p(x) = x³ + 4x² + 6x - 36
a. Through the graph, we can see that 2 is a real zero of the polynomial p. We can also use the Rational Roots Test.
p(2) = 2³ + 4.2² + 6.2 - 36 = 8 + 16 + 12 - 36 = 0
b. Now, we can use Briott-Ruffini to find the other roots and write p as a product of linear factors.
2 | 1 4 6 -36
1 6 18 0
x² + 6x + 18 = 0
Δ = 6² - 4.1.18 = 36 - 72 = -36 = 36i²
√Δ = 6i
x = -6±6i/2 = 2(-3±3i)/2
x' = -3-3i
x" = -3+3i
p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)
c. the solutions are 2, -3-3i and -3+3i
Step-by-step explanation:
The data below is what was provided in the question and it is what I solved the question with
P(A1) = 0.23
P(A2) = 0.25
P(A3) = 0.29
P(A1 n A2 ) = 0.09
P(A1 n A3) = 0.11
P(A2 n A3) = 0.07
P(A1 n A2 n A3) = 0.02
a
P(A2|A1) = P(A1 n A2)/P(A1)
= 0.09/0.23
= 0.3913
We have 39.13% confidence that event A2 will occur given that event A1 already occured
b.)
P(A3 n A3|A1) = P(A2 n A3)n A1)/P(A1)
= 0.02/0.23
= 0.08695
We have about 8.7% chance of events A2 and A3 occuring given that A1 already occured.
C.
P(A2 u A3|A1)
= P(A1 n A2)u(A1 n A3)/P(A1)
= P( A1 n A2) + P(A1 n A3) - P(A1 n A2 n A3) / P(A1)
= (0.09+0.11-0.02)/0.23
= 0.18/0.23
= 0.7826
We have 78.26% chance of A2 or A3 happening given that A1 has already occured.
Answer:
Cad
Step-by-step explanation:
I'm sure is a correct answer