We'll use the Pythagorean Theorem
130^2 = side^2 +50^2
side ^ 2 = 16,900 -2,500
side ^ 2 = 14,400
side = square root (14,400)
side = 120 feet
This question not incomplete
Complete Question
The life of a semiconductor laser at a constant power is normally distributed with a mean of 7,000 hours and a standard deviation of 600 hours. If three lasers are used in a product and they are assumed to fail independently, the probability that all three are still operating after 7,000 hours is closest to? Assuming percentile = 95%
Answer:
0.125
Step-by-step explanation:
Assuming for 95%
z score for 95th percentile = 1.645
We find the Probability using z table.
P(z = 1.645) = P( x ≤ 7000)
= P(x<Z) = 0.95
After 7000 hours = P > 7000
= 1 - P(x < 7000)
= 1 - 0.95
= 0.05
If three lasers are used in a product and they are assumed to fail independently, the probability that all three are still operating after 7,000 hours is calculated as:
(P > 7000)³
(0.05)³ = 0.125
Answer:
Step-x ≤ 3
Given
2(4 + 2x) ≥ 5x + 5 ← distribute parenthesis on left side
8 + 4x ≥ 5x + 5 ( subtract 4x from both sides )
8 ≥ x + 5 ( subtract 5 from both sides )
3 ≥ x , hence
x ≤ 3by-step explanation:
Answer:
Month 8
In this month Company B's plan will pay $64,000 versus $45,000 from Company A.
Step-by-step explanation:
Start by calculating the monthly payments for both plans.
Month - Company A - Company B
1 $10,000 $500
2 $15,000 $1,000
3 $20,000 $2,000
4 $25,000 $4,000
5 $30,000 $8,000
6 $35,000 $16,000
7 $40,000 $32,000
8 $45,000 $64,000
9 $50,000 $128,000
10 $55,000 $256,000
11 $60,000 $512,000
12 $65,000 $1,024,000
13 $70,000 $2,048,000
14 $75,000 $4,096,000
15 $80,000 $8,192,000
16 $85,000 $16,384,000
17 $90,000 $32,768,000
18 $95,000 $65,536,000
19 $100,000 $131,072,000
20 $105,000 $262,144,000
21 $110,000 $524,288,000
22 $115,000 $1,048,576,000
23 $120,000 $2,097,152,000
24 $125,000 $4,194,304,000
Answer:
Step-by-step explanation:
The equation of the line through the point 
& 
can be represented by:
Making m the subject;
∴
we need to carry out the equation of the line through (0,1) and (1,2)
i.e
y - 1 = m(x - 0)
y - 1 = mx
where;
m = 1
Thus;
y - 1 = (1)x
y - 1 = x ---- (1)
The equation of the line through (1,2) & (4,1) is:
y -2 = m (x - 1)
where;
∴
-3(y-2) = x - 1
-3y + 6 = x - 1
x = -3y + 7
Thus: for equation of two lines
x = y - 1
x = -3y + 7
i.e.
y - 1 = -3y + 7
y + 3y = 1 + 7
4y = 8
y = 2
Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7
∴
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