Answer:
a) 658008 samples
b) 274050 samples
c) 515502 samples
Step-by-step explanation:
a) How many ways sample of 5 each can be selected from 40 is just a combination problem since the order of selection isn't important.
So, the number of samples = ⁴⁰C₅ = 658008 samples
b) How many samples of 5 contain exactly one nonconforming chip?
There are 10 nonconforming chips in the batch, and 1 nonconforming chip for the sample of 5 be picked from ten in the following number of ways
¹⁰C₁ = 10 ways
then the remaining 4 conforming chips in a sample of 5 can be picked from the remaining 30 total conforming chips in the following number of ways
³⁰C₄ = 27405 ways
So, total number of samples containing exactly 1 nonconforming chip in a sample of 5 = 10 × 27405 = 274050 samples
c) How many samples of 5 contain at least one nonconforming chip?
The number of samples of 5 that contain at least one nonconforming chip = (Total number of samples) - (Number of samples with no nonconforming chip in them)
Number of samples with no nonconforming chip in them = ³⁰C₅ = 142506 samples
Total number of samples = 658008
The number of samples of 5 that contain at least one nonconforming chip = 658008 - 142506 = 515502 samples
To find the gradient of a line we first hv to find the formula
so Gradient=√X1-x2and √y1-y2
((-2,1) and (0,-5)) line 1
line 2(0,-1) and (1,0)
first we will work for line 1
X1 is -2
X2 is -0
Y1 is 1
Y2 is -1
so y=x-1
so we substitute the values
1--1=-2--0
so,
1+1=-2+0
2=-2
-0
So the equation 1 line 1 is not true
we will now go to equation 2 line 2
the equation is 3x+y=-5
we substitute the values
3(-2--0)+1--1=-5
3(-2)+-2=-5
-6+-2=-5
-8= -5
group like terms
Ur answer should look like this -3
So the answer is 3x+y=-5
your answer for this question is y=c hope this helps
Answer:
The values of 
so that 
have vertical asymptotes are 
, 
, 
, 
, 
.
Step-by-step explanation:
The function cosecant is the reciprocal of the function sine and vertical asymptotes are located at values of so that function cosecant becomes undefined, that is, when function sine is zero, whose periodicity is . Then, the vertical asymptotes associated with function cosecant are located in the values of of the form:
, 
In other words, the values of so that have vertical asymptotes are , , , , .
A)
To be similar triangles have to have equal angles
triangle ZDB'
1)angle Z=90 degrees
triangle B'CQ
1) angle C 90 degrees
angle A'B'Q=90
DB'Z+A'B'Q+CB'Q=180, straight angle
DB'Z+90+CB'Q=180
DB'Z+CB'Q=90
triangle ZDB'
DZB'+DB'Z=180-90=90
DB'Z+CB'Q=90
DZB'+DB'Z=90
DB'Z+CB'Q=DZB'+DB'Z
2)CB'Q=DZB' (these angles from two triangles ZDB' and B'CQ )
3)so,angles DB'Z and B'QC are going to be equal because of sum of three angles in triangles =180 degrees and 2 angles already equal.
so this triangles are similar by tree angles
b)
B'C:B'D=3:4
B'D:DZ=3:2
CQ-?
DC=AB=21
DC=B'C+B'D (3+4= 7 parts)
21/7=3
B'C=3*3=9
B'D=3*4=12
B'D:DZ=3:2
12:DZ=3:2
DZ=12*2/3=8
B'D:DZ=CQ:B'C
3:2=CQ:9
CQ=3*9/2=27/2
c)
BC=BQ+QC=B'Q+QC
BQ' can be found by pythagorean theorem
