Answer:
30.5 ; 49 ; 27 ; 50 ; 45
Step-by-step explanation:
Determine the first and third quartiles. Determine the second decile and the eighth decile. Determine the 67th percentile
Given the ordered data :
13 13 13 20 26 27 29 32 34 34 35 35 36 37 38 41 41 41 42 44 46 47 48 50 53 55 56 62 67 82
Sample size, n = 30
The first quartile ;Q1
Q1 = 1/4(n+1)th term
Q1 = 31/4 = 7.75th term
Q1 = (29+32)/2
Q1 = 30.5
Q3 ;
Q3 = 3/4(n+1)th term
Q3 = 3(31)/4 = 23.25 th term
Q3 = (48 + 50) /2 = 49
2nd decile :
0.2 * (nth term
0.2 * 30 = 6th term = 27
8th decile :
0.8 * 30 = 24 th term
= 50
67th percentile :
0.67 * (n+1)th term
0.67 * 31 = 20.77
= (44 + 46) / 2
= 45
A=abby's shirts
b=bik's shirts
c=cari's shirts
d=dawn's shirts
e=ellen's shirts
a> everybody
b=2.5c
d=(1/3)c
a+b+c+d=120
e=2d+1
e=13
oook
e=13=2d+1
13=2d+1
12=2d
d=6
d=(1/3)c
6=(1/3)c
18=c
b=2.5c
b=2.5(18)
b=45
a+b+c+d=120
a+45+18+6=120
a+69=120
a=51
Answer:
(A)24 cm2
Step-by-step explanation:
Answer:
The answer to your question is: 30°
Step-by-step explanation:
Data
m∠ 1 = 5x
m∠ 3 = x - 24
Process
m ∠ 1 = m ∠ 3 because they are vertical angles
5x = x - 24
5x - x = 24
4x = 24
x = 24 / 4
x = 6
Angle 1 = 5(6)
= 30°
Answer: the probability that a randomly selected Canadian baby is a large baby is 0.19
Step-by-step explanation:
Since the birth weights of babies born in Canada is assumed to be normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = birth weights of babies
µ = mean weight
σ = standard deviation
From the information given,
µ = 3500 grams
σ = 560 grams
We want to find the probability or that a randomly selected Canadian baby is a large baby(weighs more than 4000 grams). It is expressed as
P(x > 4000) = 1 - P(x ≤ 4000)
For x = 4000,
z = (4000 - 3500)/560 = 0.89
Looking at the normal distribution table, the probability corresponding to the z score is 0.81
P(x > 4000) = 1 - 0.81 = 0.19
