Answer:
the middle 50% of most lengths of pregnancies ranges between 255.62 days and 274.38 days
Step-by-step explanation:
Given that :
Mean = 265
standard deviation = 14
The formula for calculating the z score is 
x = μ + σz
At middle of 50% i.e 0.50
The critical value for 
From standard normal table
+ 0.67 or -0.67
So; when z = -0.67
x = μ + σz
x = 265 + 14(-0.67)
x = 265 -9.38
x = 255.62
when z = +0.67
x = μ + σz
x = 265 + 14 (0.67)
x = 265 + 9.38
x = 274.38
the middle 50% of most lengths of pregnancies ranges between 255.62 days and 274.38 days
Answer:
Dilation of 1/2 and then a reflection over y=2
Answer:
Exact dimensions:
Approximate dimensions:
Step-by-step explanation:
Let's assume width of rectangle is w ft
The length of a rectangular flower bed is 2ft longer than the width
so,
length =w+2
now, we can find area
now, we can plug it
we are given area =6
so, we can set it equal
and then we can solve for w
we can use quadratic formula
now, we can compare and find a,b and c
a=1 , b=2 , c=-6
we know that dimension can never be negative
so, we will only consider positive value
Exact dimensions:
Approximate dimensions:
Answer:
36.875 inches
Step-by-step explanation:
When you have 3 choices for each of 6 spins, the number of possible "words" is
3^6 = 729
The number of permutations of 6 things that are 3 groups of 2 is
6!/(2!×2!×2!) = 720/8 = 90
A) The probability of a word containing two of each of the letters is 90/729 = 10/81
The number of permutations of 6 things from two groups of different sizes is
(2 and 4) : 6!/(2!×4!) = 15
(3 and 3) : 6!/(3!×3!) = 20
(4 and 2) : 15
(5 and 1) : 6
(6 and 0) : 1
B) The number of ways there can be at least 2 "a"s and no "b"s is
15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.
_____
These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
