The given function is
f(x) = 4x - 3/2
where
f(x) = number of assignments completed
x = number of weeks required to complete the assignments
We want to find f⁻¹ (30) as an estimate of the number of weeks required to complete 30 assignments.
The procedure is as follows:
1. Set y = f(x)
y = 4x - 3/2
2. Exchange x and y
x = 4y - 3/2
3. Solve for y
4y = x + 3/2
y = (x +3/2)/4
4. Set y equal to f⁻¹ (x)
f⁻¹ (x) = (x + 3/2)/4
5. Find f⁻¹ (30)
f⁻¹ (30) = (30 + 3/2)/4 = 63/8 = 8 (approxmately)
Answer:
Pedro needs about 8 weeks to complete 30 assignments.
Answer:
Fernando incorrectly found the product of –2 and –5.
Step-by-step explanation:
Fernando evaluated the numerator of the fraction incorrectly.
Fernando simplified StartFraction 20 over 2 EndFraction incorrectly.
Fernando incorrectly found the product of –2 and –5.
Fernando evaluated (negative 3) squared incorrectly.
Fernando's calculation
5(9-5) / 2 + (-2)(-5) + (-3)^2
= 5(4) / 2 - 10 + 9
= 20/2 - 10 + 9
= 10 - 10 + 9
= 9
Correct calculation
5(9-5) / 2 + (-2)(-5) + (-3)^2
= 5(4) / 2 + (10) + 9
= 20/2 + 10 + 9
= 10 + 10 + 9
= 29
Therefore,
Fernando's error was multiplying (-2)(-5) to be equal to -10 instead of 10
Fernando incorrectly found the product of –2 and –5.
Answer:
10:1:5
Step-by-step explanation:
Given that,
A game was played by Amy, Bill, and Cara.
Let Amy's score be x. Then,
Bill's score would be x/10
Cara's score = x/2
Now,
The ratio of their scores;
x: x/10: x/2
= 1 : 1/10 : 1/2
To equalize the scores, multiply all of them by 10
= (1 * 10): (1/10 * 10) : (1/2 * 10)
= 10 : 1 : 5
Thus, the simplest ratio form would be 10:1:5
Answer:
a) 0.82
b) 0.18
Step-by-step explanation:
We are given that
P(F)=0.69
P(R)=0.42
P(F and R)=0.29.
a)
P(course has a final exam or a research paper)=P(F or R)=?
P(F or R)=P(F)+P(R)- P(F and R)
P(F or R)=0.69+0.42-0.29
P(F or R)=1.11-0.29
P(F or R)=0.82.
Thus, the the probability that a course has a final exam or a research paper is 0.82.
b)
P( NEITHER of two requirements)=P(F' and R')=?
According to De Morgan's law
P(A' and B')=[P(A or B)]'
P(A' and B')=1-P(A or B)
P(A' and B')=1-0.82
P(A' and B')=0.18
Thus, the probability that a course has NEITHER of these two requirements is 0.18.
