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tamaranim1 [39]
2 years ago
7

The expressions (30-2)(30+2) and 30^2-2^2 are equivalent and can help us find the product of two numbers. which two numbers are

they?
Mathematics
1 answer:
gtnhenbr [62]2 years ago
3 0

Answer:

  28 and 32

Step-by-step explanation:

The product indicated in your problem statement is ...

  (30 -2)(30 +2)

  = 28 × 32

The two numbers are 28 and 32.

__

The equivalent (30^2 -2^2) means that product is 900 -4 = 896.

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What are the zeros of the polynomial function y=2x^3-7x^2+2x+3?
zlopas [31]

y = 2 {x}^{3}  - 7 {x}^{2}  + 2x + 3

By inspection 1 is a root of y ,
Hence , (x-1) is a factor of y ,

\dfrac{2 {x}^{3} - 7 {x}^{2}  + 2x + 3 }{x - 1}  = 2 {x}^{2}  - 5x - 3 \\  \\ 2 {x}^{2}  - 5x - 3 = (2x + 1)(x - 3)

Hence , the 2 other roots are -1/2 and 3 , and the above graphs confirms the same.

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3 years ago
Solve the following exponential equation:<br><br>5^x=2 <br><br>e^x=10​
olya-2409 [2.1K]

Number 1.

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2 years ago
An electronic product contains 48 integrated circuits. The probability that any integrated circuit is defective is 0.01, and the
BigorU [14]

Answer:

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Step-by-step explanation:

For each integrated circuit, there are only two possible outcomes. Either they are defective, or they are not. The integrated circuits are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

An electronic product contains 48 integrated circuits.

This means that n = 48

The probability that any integrated circuit is defective is 0.01.

This means that p = 0.01

The product operates only if there are no defective integrated circuits. What is the probability that the product operates?

This is P(X = 0). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{48,0}.(0.01)^{0}.(0.99)^{48} = 0.6173

0.6173 = 61.73% probability that the product operates.

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