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Neko [114]
3 years ago
12

If n=4x−3, then what is the value of x when n=11?

Mathematics
1 answer:
crimeas [40]3 years ago
5 0

Answer:

x=7/2

Step-by-step explanation:

11=4x−3

Step 1: Flip the equation.

4x−3=11

Step 2: Add 3 to both sides.

4x−3+3=11+3

4x=14

Step 3: Divide both sides by 4.

4x /4 = 14 /4

x= 7/ 2

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Write a reclusive rule for the sequence, then give the next two numbers in the sequence 6,1,7,8,15,23
vlabodo [156]

Answer:

aₙ = aₙ₋₁ + aₙ₋₂

where:

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Step-by-step explanation:

We have the sequence:

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Is easy to see that this sequence works as follows:

You take 2 consecutive numbers, and the next number is the sum of these two.

Then we have:

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This is written as:

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A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
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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.

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