Find the common difference of the sequence shown. 1/2, 1/4, 0, ... -1/8 -1/4 -1/2

1 answer:

8 0

Common difference in Arithmetic sequences is defined as the fixed amount referring to the fact that the difference between two successive terms yields the constant value that was added. In order to know the common difference of the given sequence above, we just have to deduct 1/2 from 1/4 and the answer is -1/4. Therefore, the correct answer would be the third option. Hope this answer helps.

You might be interested in

Which answer is the same as moving the decimal in 36.2 four places to the left?

sergij07 [2.7K]

Answer:

The answer is B.dividing 36.2 by ten thousands. Plz mark as the brainlyest answer.

Step-by-step explanation:

6 0

3 years ago

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 X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X 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

I'm not sure what AAS,SAS,SSS,ASA mean .how do I solve this?

notka56 [123]

Answer:

AAS is an acronym for Angle-Angle-Side. It basically means that if two angles and a nonincluded side of one triangle are congruent to two angles and the corresponding nonincluded side of a second triangle, then the two triangles are congruent. SAS is an acronym for Side-Angle-Side. It means that if two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the two triangles are congruent. SSS is an acronym for Side-Side-Side. It means that if three sides of one triangle are congruent to three sides of a second triangle, then the two triangles are congruent. ASA is an acronym for Angle-Side-Angle. It means that if two angles and the included side of one triangle are congruent to two angles and the included side of a second triangle, then the two triangles are congruent.

In the problem, we know that the corresponding sides of both triangles are congruent to each other, so those would be given. The third side of each triangle would also be congruent because of reflexive property. Reflexive property means that the two triangles share a line segment. So, the answer would be SSS.

8 0

3 years ago

1/6 + 3/8 A. 4/6 B. 4/8 C. 4/14 D. 13/24

Naddika [18.5K]

Here is you're answer:

In order to get you're answer you need to find the common denominator then add.