Answer:
The median number of days absent is zero (0)
The mean number of days absent is 1.0 day
(b) The proportion of the population that has absenteeism greater than 4 days is 6.34 %
Step-by-step explanation:
total number of students, n = 284
The total number of students is even, the median number of days absent will in (n/2).
n/2 = 284/2 = 142
The cumulative frequency that falls in 148 students = 0 day
The median number of days absent is zero (0)
For mean:
Let the days absent = x
let the number of students = f
(b) the number of students with absenteeism greater than 4 dyas;
= 7 + 8 + 2 + 1
= 18
The proportion of these students;
Answer:
a) k
= 3.6 N/m ; b) 9.43 * 10¹²; c) 1.18 * 10¹¹; d) 75.08 N/m
Step-by-step explanation:
Length of iron bar = L = 2.7m;
side length of cross section = a = 0.07cm = 0.0007m;
x = 2.7 cm = 0.027m;
m = 100kg;
ρ = 7.87gm/cm³;
da = 2.28 * 10⁻¹⁰m;
a)
Fnet = F - mg
where Fnet = 0
So,
F = mg where F=kx
k = mg/x = 100*9.8/0.027
k = 3.6 N/m
b)
Nchain = Aw/Aa =a²/(da)²
= (o.ooo7)²/(2.28 * 10⁻¹⁰)²
= 9.43 * 10¹²
c)
Nbond = L/da = 2.7/2.28 * 10⁻¹⁰
= 1.18 * 10¹¹
d)
Spring stiffness of wire = ksi = (Nbondk)/Nchain
= [(1.18* 10¹¹)(6*10⁴)]/(9.43 * 10¹²)
= 75.08 N/m
Answer:
x=7 and y=-1
Step-by-step explanation:
X+Y=6 OR X=6-Y ...(1)
X-Y=8 ...(2)
substitue X=6-Y in (2)
(6-Y)-Y=8
6-2Y=8
-2Y=8-6
-2Y=2
Y=2/-2\Y=-1 ANS.
for x, substitute Y=-1 in (1) above
X-(-1)=8
X=8-1
X=7 ANS.
Answer:
what grade is this
Step-by-step explanation:
I'm very confused is this algebra?
