Answer:
What's a Solution to a System of Linear Equations? Note: If you have a system of equations that contains two equations with the same two unknown variables, then the solution to that system is the ordered pair that makes both equations true at the same time.
Step-by-step explanation:
Infinitely many ways!
Suppose you have the fraction 2/d.
<span>Pick </span>any<span> pair of integers a and b where b ≠ 0.</span>
Then 2b-ad is and integer, as is bd so that (2b - ad)/bd is a fraction.
Consider the fractions a/b and (2b - ad)/bd
<span>Their sum is </span>
a/b + (2b-ad)/bd = ad/bd + (2b-ad)/bd = 2b/bd = 2/d - as required.
<span>Since a and b were chosen arbitrarily, there are infinitely many possible answers to the question.</span>
Answer:
The interval from the sample of size 400 will be approximately <u>One -half as wide</u> as the interval from the sample of size 100
Step-by-step explanation:
From the question we are told the confidence level is 95% , hence the level of significance is
=> 
Generally from the normal distribution table the critical value of 
is
Generally the 95% confidence interval is dependent on the value of the margin of error at a constant sample mean or sample proportion
Generally the margin of error is mathematically represented as
Here assume that 
is constant so
=> 
=> 
So let 
and 
=> 
=> 
=> 
So From this we see that the confidence interval for a sample size of 400 will be half that with a sample size of 100
Answer:
The radius of the circle P = 2√10 = 6.325
Step-by-step explanation:
∵ AB is a tangent to circle P at A
∴ (AB)² = BC × BE
∵ BC = 8 , AB = 12 , ED = 6
∵ BE = ED + DC + CB
∴ BE = 6 + CD + 8 = 14 + CD
∴ (12)² = 8 × (14 + DC) ⇒ (12)²/8 = 14 + CD ⇒ CD = (12)²/8 - 14
∴ CD = 4
Join PC and PE (radii)
In ΔBDC and ΔPDE ⇒ ∵ ∠PDC = Ф , ∴ ∠PDE = 180 - Ф
Use cos Rule:
∵ r² = (PD)² + (DC)² - 2(PD)(DC)cosФ
∴ r² = 16 + 16 - 32cosФ = 32 - 32cosФ ⇒ (1)
∵ r² = (PD)² + (DE)² - 2(PD)(DE)cos(180 - Ф) ⇒ cos(180 - Ф) = -cosФ
∴ r² = 16 + 36 + 48cosФ = 52 + 48cosФ ⇒ (2)
∵ (1) = (2)
∴ 32 - 32 cosФ = 52 + 48cosФ
∴ 32 - 52 = 48cosФ + 32cosФ
∴ -20 = 80cosФ
∴ cosФ = -20/80 = -1/4
∴ r² = 32 - 32(-1/4) = 32 + 8 = 40
∴ r = √40 = 2√10 = 6.325
Solutions i found was m=4 or m=-2<span>
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