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3 years ago

The expression when b=3 and y= -35b-y

Mathematics

1 answer:

jekas [21]3 years ago

3 0

Answer:

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

<em>b</em> = 3

<em>y</em> = -3

5b - y

<u>Step 2: Evaluate</u>

Substitute in variables: 5(3) - -3
Multiply: 15 - - 3
Subtract: 18

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2 (r+6) + 3x +4=<br> + Scratchpad<br> Is it 5x+16

Dmitriy789 [7]

Answer:

Step-by-step explanation:

Distribute the 2(r+6)

2r + 12 + 3x + 4

Combine like terms

2r + 3x + 16

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4 years ago

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

3 years ago

what is the mean round to the nearest tenth if needed value 1 2 3 4 5 frequency 7 1 3 2 2 please help?

Blizzard [7]

1 * 7 + 2 * 1 + 3 * 3 + 4 * 2 + 5 * 2 =

7 + 2 + 9 + 8 + 10 =

Now we need to divide 36 by the total frequency: 7+1+3+2+2=15

36 / 15 = 2.4

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4 years ago

For what values of a and b are x+1 and x+2 factors of x3 – ax2 – bx –8? Write down the other factor.

emmainna [20.7K]

<h2>For a = 1 and b = 10 x+1 and x+2 factors of x³-ax²-bx-8 = 0</h2><h2>Other factor is (x-4)</h2>

Step-by-step explanation:

We have

x³-ax²-bx-8 = 0

Its factors are x+1 and x+2

That is x = -1 and x = -2 are its roots

Substituting x = -1

(-1)³-a(-1)²-b(-1)-8 = 0

-1 - a + b - 8 = 0

b - a = 9 ---------------------eqn 1

Substituting x = -2

(-2)³-a(-2)²-b(-2)-8 = 0

-8 - 4a + 2b - 8 = 0

2b - 4a = 16 ---------------------eqn 2

eqn 1 x -2

-2b + 2a = -18 ---------------------eqn 3

eqn 2 + eqn 3

-2a = -2

a = 1

Substituting in eqn 1

b - 1 = 9

b = 10

For a = 1 and b = 10, x+1 and x+2 factors of x³-ax²-bx-8 = 0

The equation is x³-x²-10x-8 = 0

Dividing with x + 1 we will get

x³-x²-10x-8 = (x+1)(x²-2x-8)

Dividing (x²-2x-8) with x + 2 we will get

x²-2x-8 = (x+2)(x-4)

So we have

x³-x²-10x-8 = (x+1)(x+2)(x-4)

Other factor is (x-4)

8 0

3 years ago

Which of the values will NOT make 3x < 45 true?

Evgesh-ka [11]

Answer:

Step-by-step explanation:

3 times 16 is 48

hope this helps :)

6 0

3 years ago

The expression when b=3 and y= -35b-y​

The expression when b=3 and y= -35b-y