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2 years ago

1)Evaluate 2A - 3B for A = 7 and B = -2.

1 answer:

5 0

Hey Brion,

1) Evaluate 2A - 3B for A = 7 and B = -2.
To solve this start by filling in the variables.
2(7) - 3(-2)

Now multiply, remember that a positive times a negative is a negative.
14 - (-6)

Finally, subtract. Remember, Keep, Change, Change (KCC). Keep the first number, change symbol (for subtraction use addition), and change the negative number to a positive number.
14 + 6 = 20

2) If x = -3, then x 2-7x + 10 equals
Start by filling in the variables.
-3(2) - 7(-3) + 10

Now follow PEMDAS (Parentheses, Exponents, Multiplication & Division, Addition & Subtraction) to solve. Remember K.C.C (no. 1 above)
-3(2) - 7(-3) + 10
(-6) - (-21) + 10
(-6) + (21) + 10
15 + 10
25*****See NOTE!***
_____________________________________________________________
Note: For #2 you may be missing a symbol between the first "x" and "2" because the answer I got above (25) is not one of your choices. Let me know what symbol is missing in the equation: x 2-7x + 10, and I will help you solve it.

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A statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20. Use thi

rjkz [21]

Answer:

\mu = 14.5\\

\sigma = 5.071\\

k = 1.084

Step-by-step explanation:

given that a statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20.

i.e. his findings with respect to probability are

$P(9$

Recall Chebyshev's inequality that

$P(|X-\mu |\geq k\sigma )\leq {\frac {1}{k^{2}}}\\P(|X-\mu |\leq k\sigma )\geq 1-{\frac {1}{k^{2}}}\\$

Comparing with the Ii equation which is appropriate here we find that

$\mu =14.5$

Next what we find is

$k\sigma = 5.5\\1-\frac{1}{k^2} =0.15\\\frac{1}{k^2}=0.85\\k=1.084\\1.084 (\sigma) = 5.5\\\sigma = 5.071$

Thus from the given information we find that

$\mu = 14.5\\\sigma = 5.071\\k = 1.084$

5 0

3 years ago

HELP ASAP!!!

Umnica [9.8K]

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

(a)/(a^2-16)+(2/(a-4))-(2/(a+4))=0

Simplify ————— a + 4 Equation at the end of step 1 : a 2 2 (—————————+—————)-——— = 0 ((a2)-16) (a-4) a+4 Step 2 : 2 Simplify ————— a - 4 Equation at the end of step 2 : a 2 2 (—————————+———)-——— = 0 ((a2)-16) a-4 a+4 Step 3 : a Simplify ——————— a2 - 16 Trying to factor as a Difference of Squares :

3.1 Factoring: a2 - 16

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =
 A2 - AB + BA - B2 =
 A2 - AB + AB - B2 =
 A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 16 is the square of 4
Check : a2 is the square of a1 

Factorization is : (a + 4) • (a - 4)

Equation at the end of step 3 : a 2 2 (————————————————— + —————) - ————— = 0 (a + 4) • (a - 4) a - 4 a + 4 Step 4 :Calculating the Least Common Multiple :

4.1 Find the Least Common Multiple

 The left denominator is : (a+4) • (a-4)

 The right denominator is : a-4

Number of times each Algebraic Factor
 appears in the factorization of: Algebraic
 Factor Left
Denominator Right
Denominator L.C.M = Max
{Left,Right} a+4 101 a-4 111

Least Common Multiple:
(a+4) • (a-4)

Calculating Multipliers :

4.2 Calculate multipliers for the two fractions

 Denote the Least Common Multiple by L.C.M
 Denote the Left Multiplier by Left_M
 Denote the Right Multiplier by Right_M
 Denote the Left Deniminator by L_Deno
 Denote the Right Multiplier by R_Deno

 Left_M = L.C.M / L_Deno = 1

 Right_M = L.C.M / R_Deno = a+4

Making Equivalent Fractions :

4.3 Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

L. Mult. • L. Num. a —————————————————— = ————————————— L.C.M (a+4) • (a-4) R. Mult. • R. Num. 2 • (a+4) —————————————————— = ————————————— L.C.M (a+4) • (a-4) Adding fractions that have a common denominator :

4.4 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

a + 2 • (a+4) 3a + 8 ————————————— = ————————————————— (a+4) • (a-4) (a + 4) • (a - 4) Equation at the end of step 4 : (3a + 8) 2 ————————————————— - ————— = 0 (a + 4) • (a - 4) a + 4 Step 5 :Calculating the Least Common Multiple :

5.1 Find the Least Common Multiple

 The left denominator is : (a+4) • (a-4)

 The right denominator is : a+4

Number of times each Algebraic Factor
 appears in the factorization of: Algebraic
 Factor Left
Denominator Right
Denominator L.C.M = Max
{Left,Right} a+4 111 a-4 101

Least Common Multiple:
(a+4) • (a-4)

Calculating Multipliers :

5.2 Calculate multipliers for the two fractions

 Denote the Least Common Multiple by L.C.M
 Denote the Left Multiplier by Left_M
 Denote the Right Multiplier by Right_M
 Denote the Left Deniminator by L_Deno
 Denote the Right Multiplier by R_Deno

 Left_M = L.C.M / L_Deno = 1

 Right_M = L.C.M / R_Deno = a-4

Making Equivalent Fractions :

5.3 Rewrite the two fractions into equivalent fractions

L. Mult. • L. Num. (3a+8) —————————————————— = ————————————— L.C.M (a+4) • (a-4) R. Mult. • R. Num. 2 • (a-4) —————————————————— = ————————————— L.C.M (a+4) • (a-4) Adding fractions that have a common denominator :

5.4 Adding up the two equivalent fractions

(3a+8) - (2 • (a-4)) a + 16 ———————————————————— = ————————————————— (a+4) • (a-4) (a + 4) • (a - 4) Equation at the end of step 5 : a + 16 ————————————————— = 0 (a + 4) • (a - 4) Step 6 :When a fraction equals zero : 6.1 When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

a+16 ——————————— • (a+4)•(a-4) = 0 • (a+4)•(a-4) (a+4)•(a-4)

Now, on the left hand side, the (a+4) • (a-4) cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
a+16 = 0

Solving a Single Variable Equation :

6.2 Solve : a+16 = 0

Subtract 16 from both sides of the equation :
 a = -16

One solution was found :

a = -16

4 0

3 years ago