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3 years ago

Find the common difference for the sequence below. 1/4, 5/16, 3/8,..

Mathematics

1 answer:

vredina [299]3 years ago

5 0

Answer:

the common difference of the sequence is $(\frac{1}{16})$

Step-by-step explanation:

A sequence of numbers is said to be in arithmetic progression of each term of the sequence has same common difference.

Now, here first term a1 = 1/4

Second term a2 = 5/16

Third term a3 = 3/8

Now, Common difference (d) = a2 - a1 = a3 - a2

Now, a2 - a1 = $\frac{5}{16} - \frac{1}{4} = \frac{5 - 4}{16} = \frac{1}{16}$

Now, a3 - a2 = $\frac{3}{8} - \frac{5}{16} = \frac{6 - 5}{16} = \frac{1}{16}$

Hence, the common difference of the sequence is $(\frac{1}{16})$

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a jet climbs at a steady rate at an angle of inclination of 0.5 degrees during takeoff. What will its height be after a 2.2km in

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2 years ago

Given: 8x-61=17, prove: 39/4

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Answer:

The solution is $x = \frac{39}{4}$

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We are given the following equation:

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3 years ago

2.3 x 10^4 x 6.7 x 10^3 divide by <br> 5 x 10-^8

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3 years ago

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper

Korvikt [17]

Answer:

90.67% probability that John finds less than 7 golden sheets of paper

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it contains a golden sheet of paper, or it does not. The probability of a container containing a golden sheet of paper is independent of other containers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper containers.

This means that $p = 0.3$

14 of these containers of paper.

This means that $n = 14$

What is the probability that John finds less than 7 golden sheets of paper?

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.3)^{0}.(0.7)^{14} = 0.0068$

$P(X = 1) = C_{14,1}.(0.3)^{1}.(0.7)^{13} = 0.0407$

$P(X = 2) = C_{14,2}.(0.3)^{2}.(0.7)^{12} = 0.1134$

$P(X = 3) = C_{14,3}.(0.3)^{3}.(0.7)^{11} = 0.1943$

$P(X = 4) = C_{14,4}.(0.3)^{4}.(0.7)^{10} = 0.2290$

$P(X = 5) = C_{14,5}.(0.3)^{5}.(0.7)^{9} = 0.1963$

$P(X = 6) = C_{14,6}.(0.3)^{6}.(0.7)^{8} = 0.1262$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0068 + 0.0407 + 0.1134 + 0.1943 + 0.2290 + 0.1963 + 0.1262 = 0.9067$

90.67% probability that John finds less than 7 golden sheets of paper

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2 years ago