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3 years ago

Kiran's father is 65 years old. He is 8 years older than three

Mathematics

2 answers:

I am Lyosha [343]3 years ago

8 0

Answer:

kiran is 19 years old.

Step-by-step explanation:

kiran's father age = 65

let kiran's age be x

kiran's father is 8 years older than three times kiran's age means 8 + 3x

so according to the question equation is

65 = 8 + 3x

65 - 8 = 3x

57/3 = x

19 = x

therefore kiran's age is 19 years.

natka813 [3]3 years ago

4 0

Answer:

let kiran's age be x

kiran's father's age is 65

acc.to ques

kiran's father is 8 years older than 3 times kiran's age ....which gives us the equation as follows

8+3x =65

therefore

3x =65-8

3x =57

x =57/3

x= 19 ans

so kiran's age is 19 years.

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ki77a [65]

Answer:

<h2>no solution, the equation is false</h2>

Step-by-step explanation:

what is the solution for the following equation 3(2x+3)=2(3x+4)?

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2 years ago

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Please help me with this question!!

stepladder [879]

Answer:

rational
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irrational
irrational
√7, it is irrational

Step-by-step explanation:

A rational number is one that can be expressed as the ratio of two integers. All fractions that have integer numerators and (non-zero) denominators are rational numbers. Any finite decimal number, or any repeating decimal number, is a rational number. These can always be expressed as the ratio of two integers. For example, 0.4040... = 40/99, and 0.286 = 286/1000.

To make an irrational sum, at least one of the contributors must be irrational. You want an irrational 2-number sum that has 7/8 as one of the contributors. Since 7/8 is rational, the other contributor must be irrational.

Step 1. The number 7/8 is rational.

Step 2. The desired sum is irrational.

Step 3. The rule rational + irrational = irrational applies.

Step 4. An irrational number must be chosen.

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3 years ago

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

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

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

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

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

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

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

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

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

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