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ycow [4]
3 years ago
6

The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original

number, we get 143. What can be the original number?
Mathematics
2 answers:
Zanzabum3 years ago
6 0

Answer:

The original number could be 85.

Step-by-step explanation:

Let the 2 digits be x and y.

Let the number be xy then, assuming that x is the larger digit:

x - y = 3.

x = y + 3

Also

10y + x + 10x + y = 143

Substituting for x:

10y + y + 3 + 10(y + 3) + y = 143

22y + 33 = 143

22y = 110

y = 5.

So x = y + 3 = 8.

marin [14]3 years ago
3 0

Answer:

  • Let the unit digit be x and tens digit be x + 3

  • Therefore, the original number = 10(x + 3) + x

  • On interchanging, the number formed = 10x + x + 3

<h3><u>❍ According to Question now,</u></h3>

➥ 10(x + 3) + x + 10x + x + 3 = 143

➥ 10x + 30 + 12x + 3 = 143

➥ 22x + 33 = 143

➥ 22x = 143 - 33

➥ 22x = 110

➥ x = 110/22

➥ x = 5

<h2>__________________...</h2>

<h3>Therefore,</h3>

The unit digit number = x = 5

The tens digit number = x + 3 = 5 + 3 = 8

<h2>__________________...</h2>

The original number = 10(x + 3) + x

The original number = 10(5 + 3) + 5

The original number = 50 + 30 + 5

The original number = 85

Hence<u>,the original number is 85.</u>

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3 years ago
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Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

<em>Let </em>\bar X<em> = sample average time spent studying</em>

The z-score probability distribution for sample mean is given by;

          Z = \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }  ~ N(0,1)

where, \mu = population mean hours spent studying = 25 hours

            \sigma = standard deviation = 15 hours

            n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < \bar X < 30 hours)

    P(29 hours < \bar X < 30 hours) = P(\bar X < 30 hours) - P(\bar X \leq 29 hours)

      

    P(\bar X < 30 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } < \frac{ 30-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z < 2) = 0.97725

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<em>So, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.</em>

Therefore, P(29 hours < \bar X < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

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