Answer:
2587mm^3 approx!
Step-by-step explanation:
first you divide the nut into 6 part(in triangle now, by joining centre to each edge)
let's take one part of the triangular shape then area of that part can be found by using 1/2×base×height
i.e, 1/2×13×15=97.5(mm^2)
now when we consider depth of that traingular part,we will get volume of that part as area×depth
i.e, 97.5×6=585(mm^3)
now volume of all the 6 triangular part is 585×6=3510(in mm^3)
now take circular cavity in consideration, it's volume will be π(7^2)6=923(mm^3) approximately
now reqired volume will be volume of that hexagonal part minus that of circular cavity
=3510-923
=2587mm^3
✌️
Answer:
A. 0.62%
B. 28 months
Step-by-step explanation:
A. Calculation for what percentage of total production will the company expect to replace
Let x represents the distribution of life times
Let mean be 34 months
Let standard deviation be 4 months.
Based on the information the full refund on any defective watch for 2 years will represent 24 months (2 years *12 months).
First step
P(X<24)
= p(x-mean/ standard deviation< 24-34/4)
= p(z< -10/4)
=P(z<-2.5)
Second step is to Use the excel function to find NORMSDIST(z) of P(z<-2.5)
NORMSDIST(z)=0.62%
Therefore the percentage of total production will the company expect to replace will be 0.62%
B. Calculation for how much the guarantee period should be
First step
P(X<x)=0.06
P(x-Mean/Standard deviation < x-34/4) = 0.06
Second Step is to Use excel function
P(z<x-34/4) = (Normsinv(0.06)
x-34/4 = -1.555
Now let calculate how much the guarantee period should be
x = -6.22+34 months
x = 27.78
x = 28 months (Approximately)
Therefore the guarantee period should be 28 months
Answer:
15% markdown
Step-by-step explanation:
To find the percent markdown
Take the original price minus the new price
975-828.75
146.25
Divide by the original price
146.25/975
.15
Change to percent form
15% markdown
Answer:
x = -4/3
Step-by-step explanation:
6x + 11 = -6x - 5
12x = -16
x = -4/3
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ
Use the Sum/Difference Identities:
sin(α + β) = sinα · cosβ + cosα · sinβ
cos(α - β) = cosα · cosβ + sinα · sinβ
Use the Unit circle to evaluate: sin45 = cos45 = √2/2
Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ
Use the Pythagorean Identity: cos²Ф + sin²Ф = 1
<u />
<u>Proof LHS → RHS</u>
LHS: 2sin(45 + 2A) · cos(45 - 2A)
Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)
Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]
Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]
Distribute: cos²2A + 2cos2A·sin2A + sin²2A
Pythagorean Identity: 1 + 2cos2A·sin2A
Double Angle: 1 + sin4A
LHS = RHS: 1 + sin4A = 1 + sin4A 