Answer:
Step-by-step explanation:
A+N+J=310
A=N*2 =>N=A/2
N=J*3 =>J=N/3
N=?
N*2+N+N/3=310
3N+N/3=310
(9N+N)/3=930/3
10N=930
N=930/10
N=93
will be nice if you give me brainlies.Good luck!
Answer: the qualifying time in seconds is about 25.3
Step-by-step explanation:
Since the personal best finishing times for a particular race in high school track meets are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = personal best finishing times for a particular race.
µ = mean finishing time
σ = standard deviation
From the information given,
µ = 24.6 seconds
σ = 0.64 seconds
The probability value for the top 15% of finishing time for runners to qualify would be (1 - 15/100) = (1 - 0.15) = 0.85
Looking at the normal distribution table, the z score corresponding to the probability value is 1.04
Therefore,
1.04 = (x - 24.6)/0.64
Cross multiplying by 0.64, it becomes
1.04 × 0.64 = x - 24.6
0.6656 = x - 24.6
x = 0.6656 + 24.6
x = 25.3 seconds
Answer:
Step-by-step explanation:
Answer:
6 ; 9 ; 11 ; 12.5; 14
B.) 3.5
Step-by-step explanation:
Given the data:
6 9 9 10 11 11 12 13 14
a.) The low = 6 (lowest value in the dataset)
b.) Q1 = Lower quartile
Q1 = 1/4(n + 1)th term
n = sample size = 9
Q1 = 1/4(9 + 1) ; 1/4(10) = 2.5th term
Q1 = (2nd + 3rd term) / 2 = (9 + 9) / 2 = 9
Median Q2:
Q2 = 1/2(9 + 1) ; 1/2(10) = 5th term = 11
Upper Quartile Q3:
Q3 = 3/4(9 + 1) ; 3/4(10) = 7.5th term
Q3 = (7th + 8th term) / 2 = (12 + 13) / 2 = 12.5
Interquartile range (Q3 - Q1)
(12.5 - 9) = 3.5
The high = 14 (highest value in the dataset)
