Answer:
(1, 2 )
Step-by-step explanation:
Given the 2 equations
- 2x - y = - 4 → (1)
x + 2y = 5 → (2)
Multiplying (2) by 2 and adding result to (1) will eliminate x- term
2x + 4y = 10 → (3)
Add (1) and (3) term by term to eliminate x
0 + 3y = 6
3y = 6 ( divide both sides by 3 )
y = 2
Substitute y = 2 into either of the 2 equations and solve for x
Substituting into (2)
x + 2(2) = 5
x + 4 = 5 ( subtract 4 from both sides )
x = 1
solution is (1, 2 )
Answer:
The 1st table represents a function
Step-by-step explanation:
square tan theta
(tan^2) + 1 = sec^2
so sec^2 = 11/4
sec = - (sqrt 11)/2
cos= - 2/ (sqrt 11)
answer choice a if you simplify
Jack paid $ 23000 for the ring.
<u>Solution:</u>
Given that, Jack Matthew bought a new diamond ring for $20,000
Sales tax is 5% with a 10% excise tax
<em><u>To find: total price including taxes</u></em>
Given that sales tax is 5% of 20,000
![\begin{array}{l}{\text { Sales tax }=5 \% \times \text { cost price }} \\\\ {\text { Sales tax }=\frac{5}{100} \times 20000=5 \times 200=1000}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Ctext%20%7B%20Sales%20tax%20%7D%3D5%20%5C%25%20%5Ctimes%20%5Ctext%20%7B%20cost%20price%20%7D%7D%20%5C%5C%5C%5C%20%7B%5Ctext%20%7B%20Sales%20tax%20%7D%3D%5Cfrac%7B5%7D%7B100%7D%20%5Ctimes%2020000%3D5%20%5Ctimes%20200%3D1000%7D%5Cend%7Barray%7D)
Also given that excise tax is 10%
![\begin{array}{l}{\text { Excise tax }=10 \% \times \text {cost price }} \\\\ {\text { Excise tax }=\frac{10}{100} \times 20000=10 \times 200=2000}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Ctext%20%7B%20Excise%20tax%20%7D%3D10%20%5C%25%20%5Ctimes%20%5Ctext%20%7Bcost%20price%20%7D%7D%20%5C%5C%5C%5C%20%7B%5Ctext%20%7B%20Excise%20tax%20%7D%3D%5Cfrac%7B10%7D%7B100%7D%20%5Ctimes%2020000%3D10%20%5Ctimes%20200%3D2000%7D%5Cend%7Barray%7D)
<em>Total price = cost price + sales tax + excise tax</em>
Total price = 20,000 + 1000 + 2000
Total price = 23000
Hence, jack paid $ 23000 for the ring.
Answer:
A. 0.1035
B. 0.1406
C. 0.1025
Step-by-step explanation:
Given that:
the number of sample questions (n) = 5
The probability of choosing the correct choice (p) = 1/4 = 0.25
Suppose X represents the number of question that are guessed correctly.
Then, the required probability that she gets the majority of her question correctly is:
P(X>2) = P(X=3) + P(X =4) + P(X = 5)
![P(X>2) = [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4} + (^{5}C_{5}) \times (0.25)^5 (1-0.25)^{5-5}](https://tex.z-dn.net/?f=P%28X%3E2%29%20%3D%20%20%5B%20%28%5E%7B5%7DC_%7B3%7D%29%20%5Ctimes%20%280.25%29%5E3%20%281-0.25%29%5E%7B5-3%7D%20%2B%20%28%5E%7B5%7DC_%7B4%7D%29%20%5Ctimes%20%280.25%29%5E4%20%281-0.25%29%5E%7B5-4%7D%20%2B%20%28%5E%7B5%7DC_%7B5%7D%29%20%5Ctimes%20%280.25%29%5E5%20%281-0.25%29%5E%7B5-5%7D)
![P(X>2) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!} \times (0.25)^4 (1-0.25)^{1} +\dfrac{5!}{5!(5-5)!} \times (0.25)^5 (1-0.25)^{0} \Bigg ]](https://tex.z-dn.net/?f=P%28X%3E2%29%20%3D%20%5CBigg%20%5B%20%5Cdfrac%7B5%21%7D%7B3%21%285-3%29%21%7D%20%5Ctimes%20%280.25%29%5E3%20%281-0.25%29%5E%7B2%7D%20%2B%20%5Cdfrac%7B5%21%7D%7B4%21%285-4%29%21%7D%20%20%5Ctimes%20%280.25%29%5E4%20%281-0.25%29%5E%7B1%7D%20%2B%5Cdfrac%7B5%21%7D%7B5%21%285-5%29%21%7D%20%20%5Ctimes%20%280.25%29%5E5%20%281-0.25%29%5E%7B0%7D%20%5CBigg%20%5D)
P(X>2) = [ 0.0879 + 0.0146 + 0.001 ]
P(X>2) = 0.1035
B.
Recall that
n = 5 and p = 0.25
The probability that the first Q. she gets right is the third question can be computed as:
![P(X=x) = 0.25 ( 1- 025) ^{x-1}](https://tex.z-dn.net/?f=P%28X%3Dx%29%20%3D%200.25%20%28%201-%20025%29%20%5E%7Bx-1%7D)
Since, x = 3
![P(X = 3) = 0.25 ( 1- 0.25 ) ^{3-1}](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%200.25%20%28%201-%200.25%20%29%20%5E%7B3-1%7D)
![P(X =3) = 0.25 (0.75)^{3-1}](https://tex.z-dn.net/?f=P%28X%20%3D3%29%20%3D%200.25%20%280.75%29%5E%7B3-1%7D)
![P(X =3) = 0.25 (0.75)^{2}](https://tex.z-dn.net/?f=P%28X%20%3D3%29%20%3D%200.25%20%280.75%29%5E%7B2%7D)
P(X=3) = 0.1406
C.
The probability she gets exactly 3 or exactly 4 questions right is as follows:
P(X. 3 or 4) = P(X =3) + P(X =4)
![P(X=3 \ or \ 4) = [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4}]](https://tex.z-dn.net/?f=P%28X%3D3%20%5C%20or%20%5C%204%29%20%3D%20%20%5B%20%28%5E%7B5%7DC_%7B3%7D%29%20%5Ctimes%20%280.25%29%5E3%20%281-0.25%29%5E%7B5-3%7D%20%2B%20%28%5E%7B5%7DC_%7B4%7D%29%20%5Ctimes%20%280.25%29%5E4%20%281-0.25%29%5E%7B5-4%7D%5D)
![P(X=3 \ or \ 4) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!} \times (0.25)^4 (1-0.25)^{1} \Bigg ]](https://tex.z-dn.net/?f=P%28X%3D3%20%5C%20or%20%5C%204%29%20%3D%20%5CBigg%20%5B%20%5Cdfrac%7B5%21%7D%7B3%21%285-3%29%21%7D%20%5Ctimes%20%280.25%29%5E3%20%281-0.25%29%5E%7B2%7D%20%2B%20%5Cdfrac%7B5%21%7D%7B4%21%285-4%29%21%7D%20%20%5Ctimes%20%280.25%29%5E4%20%281-0.25%29%5E%7B1%7D%20%5CBigg%20%5D)
P(X = 3 or 4) = [ 0.0879 + 0.0146 ]
P(X=3 or 4) = 0.1025