Answer:
(4,2)
Step-by-step explanation:
The x coordinate of the midpoint is the sum of the x coordinates divided by 2
( 1+7)/2 = 8/2 = 4
The y coordinate of the midpoint is the sum of the y coordinates divided by 2
( 2+2)/2 = 4/2 = 2
The midpoint is (4,2)
8(6x + 5) + 2x
8(6x) + 8(5) + 2x
48x + 40 + 2x
50x + 40
Answer:
The test statistic = -0.93
Step-by-step explanation:
The test statistic is given by the formula
z = (X₁ - X₂) ÷ √(σₓ₁² + σₓ₂²)
where X₁ = proportion of data of South Korean tourists = (57/134) = 0.425
X₂ = proportion of other country tourists = (72/150) = 0.48
σₓ₁ = standard error in data 1 = √[p(1-p)/n]
= √(0.425 × 0.575/134) = 0.0427
σₓ₂ = standard error in data 2 = √[p(1-p)/n]
= √(0.48 × 0.52/150) = 0.0408
z = (X₁ - X₂) ÷ √(σₓ₁² + σₓ₂²)
z = (0.425 - 0.48) ÷ √(0.0427² + 0.0408²)
z = -0.055 ÷ 0.0590586996
z = -0.9313
Hope this Helps!!!
Answer: No, it would be 2/5.
Step-by-step explanation: 2 1/4 - 6/7 is equal to about 1.4 when rounded to the nearest tenth. 1.4 as a fraction would be 4/10. 4/10 is simplified to 2/5 by dividing each number by 2.
Answer:
The probability that an athlete chosen is either a football player or a basketball player is 56%.
Step-by-step explanation:
Let the athletes which are Football player be 'A'
Let the athletes which are Basket ball player be 'B'
Given:
Football players (A) = 13%
Basketball players (B) = 52%
Both football and basket ball players = 9%
We need to find probability that an athlete chosen is either a football player or a basketball player.
Solution:
The probability that athlete is a football player = 
The probability that athlete is a basketball player = 
The probability that athlete is both basket ball player and football player = 
We have to find the probability that an athlete chosen is either a football player or a basketball player
.
Now we know that;

Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.