Answer:
A) y² - 5y + 1
B) y² - 5y - 4
C) - 5y + 1
D) - 3y² - 4y - 6
Step-by-step explanation:
Let's call P the unkown polynomial and D the difference. In each case, the following must be true:
y² - 5y + 1 - P = D
<em>A)</em>
y² - 5y + 1 - P = 0
y² - 5y + 1 = P
<em>B)</em>
y² - 5y + 1 - P = 5
y² - 5y + 1 - 5 = P
y² - 5y - 4 = P
<em>C)</em>
y² - 5y + 1 - P = y²
y² - 5y + 1 - y² = P
- 5y + 1 = P
<em>D) </em>
y² - 5y + 1 - P = 4y² - y + 7
y² - 5y + 1 - 4y² + y - 7 = P
- 3y² - 4y - 6 = P
2/3
16/24 and divide the greatest common factor (8)
Answer:
A.6.2i-4.2 j
Step-by-step explanation:
We are given that
We have to find the projection of w on to u.
The project of vector a on b
=
Using the formula
The projection of w on to u
The projection of w on to u=6.2i-4.2j
Hence, option A is true.
Answer: the price of a senior citizen's ticket is $8.
the price of a child's ticket is $14
Step-by-step explanation:
Let x represent the price of a senior citizen's ticket.
Let y represent the price of a child's ticket.
On the first day of ticket sales, the school sold 3 senior citizen tickets and 1 child ticket for a total of $38. It means that
3x + y = 38- - - - - - - - - - - -1
The school took in $52 on the second day by selling 3 senior citizen and 2 child tickets. It means that
3x + 2y = 52- - - - - - - - - - - -2
Subtracting equation 2 from equation 1, it becomes
- y = - 14
y = 14
Substituting y = 14 into equation 1, it becomes
3x + 14 = 38
3x = 38 - 14 = 24
x = 24/3
x = 8
Answer:
Step-by-step explanation:
A) From the stem-leaf plot, we see that out of 21 tunas, 5 have dangerous levels of copper since the levels go beyond 5.7 parts per million. The required proportion is 5/21=0.2381
B) Given the sample mean is {x}=4.77, sample standard deviation s=1.16 and the sample size is n=21.
Since the population standard deviation is not known, we use t-distribution.
So the 98% CI for mean is
4.77 ± t{1-0.02 /2,20} x 1.16/sqrt(21) = (4.13, 5.41)}
We are sure with 98% confidence the true copper level (in parts per million) lies in the interval (4.13, 5.41)
