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gregori [183]
3 years ago
10

The area of a rectangle is 12x2 - 8x - 15. The width is (2x - 3). What is the length of the

Mathematics
1 answer:
olga_2 [115]3 years ago
8 0
The answer to this problem is
b. (6x +5).
Hope this helps!!;)
Brainiest please!:)
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Varvara68 [4.7K]
So hmm check the picture below

what's the width? well, w = l + 3

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3 years ago
What is the answer to: 3 x 4 + 15- 19 x2037? help
Ierofanga [76]

Answer: -38,686

Step-by-step explanation:

3 x 4= 12

-19 x 2037= -38,703

12 + 15= 17

17 -38,703= -38,686

6 0
3 years ago
What is the reciprocal
weqwewe [10]
Answer is C. 7/12 has the reciprocal of 12/7. When You multiply by a reciprocal, you just switch the numerator and denominator. The answer does NOT become negative. It would change the fraction itself.
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5/7= 7/5
3 0
3 years ago
Read 2 more answers
If 62% of 80 classmates like hip-hop, but 70% of 60 relatives like hip-hop, which one is more?
SIZIF [17.4K]

62 percent of 80 is more

Step-by-step explanation:

the 62 percent of 80 comes out to 49.6 people and the other one comes out to 42 people

i dont know how to explain how to do it

8 0
2 years ago
Read 2 more answers
A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
3 years ago
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