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ella [17]
4 years ago
11

7b squared -21b-273=7

Mathematics
2 answers:
guajiro [1.7K]4 years ago
8 0

Answer:

Given 7b²-21b-273=7, the solutions are x1 = 8 and x2 = -5.

Step-by-step explanation:

Given 7b²-21b-273=7, first you need to equal zero. So

7b²-21b-273-7=0 ⇒ 7b²-21b-280 = 0

The secon step is to find the solutions applying Bhaskara´s formula x = (-b ± √(b²-4×a×c))/2×a

Where a=7, b= -21 and c= -280

After you identified each term, you have to replace it on the formula so....

x = (21 ± √(21² - 4×7×(-280)))/2×7 ⇒ x = (21 ± √(441 + 7840))/14                   ⇒ x = (21 ± √8281)/14

Then you will obtain two values for x, called x1 = 8 and x2=-5.

makvit [3.9K]4 years ago
8 0

7b^2 -21b -273 =7

Moving the -273 to the right hand side

7b^2 -21b =7+273

7b^2 -21b =280

Dividing the whole equation by 7,thus 7b^2/7 -21b/7 = 280/7

You will get

b^2 -3b =40.

Solving the equation as quadratic by moving the 40 to left side and equating the equation to zero, thus

b^2 -3b -40 =0

Factorizing b , you will get

(b-8) and (b+5) =0 thus b=8 ,b=-5 respectively. Now substituting the -5 in the equation satisfies it. Therefore b = -5

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Answer:

Step-by-step explanation:

Hello!

In a chemical adhesive manufacturing company you can encounter the following situations:

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40% of all batches with the material of two or more lots require additional processing.

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f. You have to calculate the probability of "batch is formed with two different lots" and "the lot does not require additional processing", these two events aren't independent.

You cannot calculate the probability of this intersection the same way you did in item e. because we have no information about the probability of P(B'/A). But the P(A) is equal to the summation of P(A ∩ B) and P(A ∩ B'), so:

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First calculate P(A' ∩ B) = P(A')* P(B/A')= 0.91*0.05= 0.0455

P(B)= P(A ∩ B) + P(A' ∩ B)= 0.036 + 0.0455= 0.0815

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