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2 years ago

Find all possible values of b such that 4x² + bx + 3 can be factored. Check all that apply.

Mathematics

1 answer:

aleksklad [387]2 years ago

3 0

The possible values of b are 7 and 8

<h3>How to determine the possible values of x?</h3>

The expression is given as:

4x² + bx + 3

Next, we test the options to determine the values of b

Option 1: b = 13

So, we have:

4x² + 13x + 3 ---- this cannot be factorized

Option 2: b = 7

So, we have:

4x² + 7x + 3

Expand

4x² + 4x + 3x + 3

Factorize

4x(x +1) + 3(x + 1)

Factor out x + 1

(4x + 3)(x + 1) ------ this can be factorized

Option 3: b = 8

So, we have:

4x² + 8x + 3

Expand

4x² + 6x + 2x + 3

Factorize

2x(2x +3) + 1(2x + 3)

Factor out 2x + 3

(2x + 3)(2x + 1) ------ this can be factorized

Option 4: b = 1

So, we have:

4x² + x + 3 ---- this cannot be factorized

Hence, the possible values of b are 7 and 8

Read more about factorized expressions at:

brainly.com/question/723406

#SPJ1

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Given: ABCD is a parellelogram. Prove: AB=CD and BC=DA

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Suppose a manufacturer finds that 95% of their production is normal but the final 5% has one or more flaws. Each flawed good has

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Answer:

FLAW TYPE2 NO TYPE2 FLAW

TYPE1 0.015 0.025

NO TYPE1 FLAW 0.01 0.95

2) 0.04 and $0.04

3) 0.025 and $0.025

4) 0.015 and $0.015

5) 0.95 and $0.95

Step-by-step explanation:

Given that;

financial cost = $1

p(flaw) = 0.05

p(type 1 flaw / flaw) = 80% = 0.8

p(type 2 flaw / flaw) = 50% = 0.5

p( type 1 and 2 flaw/flaw) = 30% = 0.30

1) Bivariate Table

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw) = 0.05 × 0.5 = 0.025

p( type 1 and 2 flaw) = p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

p( only 1 flow) = 0.04 - 0.015 = 0.025

p( only 2 flow) = 0.025 - 0.015 = 0.01

THEREFORE the Bivariate Table;

FLAW TYPE2 NO TYPE2 FLAW

TYPE1 0.015 0.025

NO TYPE1 FLAW 0.01 0.95

2) probability and expectations of type 1 flaw?

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

Expected financial cost to the firm per good = $1 × 0.04 = $0.04

3) probability and expectation of Type 2 flaw

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw) = 0.05 × 0.5 = 0.025

Expected financial cost to the firm per good = $1 × 0.025 = $0.025

4) probability and expectations of Type 1 and 2 flaws

p( type 1 and 2 flaw) = p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

Expected financial cost to the firm per good = $1 * 0.015 = $0.015

5) probability and expectations of no flaws?

Probability of no flaw = P(No flaw) =95% = 0.95

Expected financial cost saved the firm per good due to no flaw

= $1 × 0.95 = $0.95

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