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Sonja [21]
3 years ago
15

What is the product ? (4y-3)(2y^2+3y-5)

Mathematics
2 answers:
DIA [1.3K]3 years ago
5 0

Answer:

=8y^3+6y^2−29y+15

Step-by-step explanation:

(4y−3)(2y^2+3y−5)

=(4y+−3)(2y^2+3y+−5)

=(4y)(2y^2)+(4y)(3y)+(4y)(−5)+(−3)(2y^2)+(−3)(3y)+(−3)(−5)

=8y^3+12y^2−20^y−6y^2−9y+15

=8y^3+6y^2−29y+15

yuradex [85]3 years ago
4 0

Answer:

8y^3 + 6y^2 - 29y + 15

Step-by-step explanation:

8y^3 + 12y^2 - 20y - 6y^2 - 9y + 15

then

8y^3 + 6y^2 - 29y + 15 is the product

Best regards

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3 years ago
A candy company claims that 20% of the candies in its bags are colored green. Steve buys 30 bags of 30 candies, randomly selects
Allisa [31]

Answer:

The probability of Steve agreeing with the company’s claim is 0.50502.

Step-by-step explanation:

Let <em>X</em> denote the number of green candies.

The probability of green candies is, <em>p</em> = 0.20.

Steve buys 30 bags of 30 candies, randomly selects one candy from each, and counts the number of green candies.

So, <em>n</em> = 30 candies are randomly selected.

All the candies are independent of each other.

The random variable <em>X</em> follows a binomial distribution with parameter <em>n</em> = 30 and <em>p</em> = 0.20.

It is provided that if there are 5, 6, or 7 green candies, Steve will conclude that the company’s claim is correct.

Compute the probability of 5, 6 and 7 green candies as follows:

P(X=5)={30\choose 5}(0.20)^{5}(1-0.20)^{30-5}=0.17228\\\\P(X=6)={30\choose 6}(0.20)^{6}(1-0.20)^{30-6}=0.17946\\\\P(X=7)={30\choose 7}(0.20)^{7}(1-0.20)^{30-7}=0.15328

Then the probability of Steve agreeing with the company’s claim is:

P (Accepting the claim) = P (X = 5) + P (X = 6) + P (X = 7)

                                       = 0.17228 + 0.17946 + 0.15328

                                       = 0.50502

Thus, the probability of Steve agreeing with the company’s claim is 0.50502.

7 0
3 years ago
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