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3 years ago

Find the sum of 7/13 and 11/26.

Mathematics

2 answers:

Angelina_Jolie [31]3 years ago

5 0

Answer:

25/26

Step-by-step explanation:

7/13 + 11/26

The common denominator is 26

7/13 *2/2 +11/26

14/26 + 11/26

25/26

timofeeve [1]3 years ago

4 0

Answer:

The sum of 7/13 and 11/26 is 25/26.

Step-by-step explanation:

First off you find the LCM (Least Common Multiple) between the denominators which are 13 and 26 in this case. The LCM between 13 and 26 would be 26. You then change 13 to the denominator of 26. 13x2 is 26 so whatever you do to one side or one part, you have to do to the other. Now, you will multiply 7x2 to get 14. Your final fractions are 14/26 and 11/26. When you add 14/26 and 11/26, you get 25/26. Hope this helps! :)

<h2>CloutAnswers</h2>

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Choose the correct sum of the polynomials (4x3 − 2x − 9) (2x3 5x 3). 6x3 − 3x − 6 2x3 − 7x − 12 6x3 3x − 6 2x3 3x − 3.

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The correct sum of the polynomial is $\rm 6x^3 + 3x - 6$ .

We have to determine

The correct sum of the polynomials.

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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

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.

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)

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