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3 years ago

What is answer of equation 1/3 + f = 5

Mathematics

2 answers:

Gala2k [10]3 years ago

8 0

Answer:

Step-by-step explanation:

1/3 + f = 5

f = 5 - 1/3

f = 15/3 - 1/3

f = 14/3 or 4 2/3 <===

aleksklad [387]3 years ago

3 0

Pretty sure the answer is F=4 2/3 I think

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The sales tax is 2.10

Step-by-step explanation:

Take the price of the item and multiply by the sales tax rate to determine the tax

30 * 7%

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15. a music concert is organized at a memorial auditorium. the first row of the auditorium has 16 seats, the second row has 24 s

Drupady [299]

The equation for the nth term in the arithmetic sequence is 8n + 8.

The number of people that can be accommodated in the 16th row is 136.

<h3>What is an arithmetic progression?</h3>

Arithmetic Progression (AP) is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value. It is also called Arithmetic Sequence.

Given that,

No. of seats in first row = 16

No. of seats in second row = 24

No. of seats in third row = 32

Total number of rows = 50

It forms an arithmetic progression

First term = a = 16

common difference d = 8

Number of terms, n = 50

(A) The formula for the n th term of an arithmetic progression is given by

Tn = a + (n - 1) d

= 16 + (n-1) 8

= 16 + 8n - 8

Tn = 8n + 8

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The number of seats in 16 th row is given by

T(16) = 8 x 16 + 8

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Hence, (A)The equation for the nth term in the arithmetic sequence is 8n + 8. and (B) The number of people that can be accommodated in the 16th row is 136.

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4 years ago

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This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi

svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that 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.

3% of the components are identified as defective

This means that $p = 0.03$

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is $P(X = 0)$ when $n = 90$ . So

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

$P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645$

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so $P(X \leq 2)$ when $n = 102$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

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

$P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447$

$P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411$

$P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062$

0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so $P(X \leq 5)$ when $n = 105$

Then

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

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

$P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408$

$P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326$

$P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133$

$P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265$

$P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786$

$P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034$

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

3 0

3 years ago