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3 years ago

The units digit of a three-digit number is 5. The sum of its digits is 11. If the units and

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

3 0

Answer:

The original number is 245.

Step-by-step explanation:

Let A be the hundreds digit of the original number, B be the tens digit of the original number, and C be the units digit of the original number.

We know from the first two sentences in the problem that:

C = 5

A + B + C = 11

Substituting the value of C, we get:

A + B + 5 = 11

A + B = 6

In the third sentence, we are told that when the number is added to the reverse of itself, the sum of the two numbers is 787. Expressed mathematically:

100A + 10B + C + A + 10B + 100C = 787

Where (100A + 10B + C) gives the value of the original number and (A + 10B + 100C) gives the value of the number when reversed. Simplified, we get:

101A + 20B + 101C = 787

Rewriting the "A + B = 6" equation from above as B = 6 - A, we substitute (6 - A) into the equation above for B and substitute 5 for C (because we know C = 5):

101A + 20(6 - A) + 101 * 5 = 787

101A + 120 - 20A + 505 = 787

81A + 625 = 787

81A = 162

A = 2

Now that we have A, we can get B using A + B = 6:

2 + B = 6

B = 4

To summarize:

A = 2

B = 4

C = 5

So the original number must be 245.

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What is the equation of the line that passes through the point (5,4) and is perpendicular to the line whose equation is 2x + y =

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The equation of the line that passes through the point (5,4) and is perpendicular to the line whose equation is 2x + y = 3 in standard form is x - 2y = - 3

Solution:

Given, line equation is 2x + y = 3

2x + y – 3 = 0 ----- eqn (1)

We have to find a line that is perpendicular to 2x + y – 3 = 0 and passing through (5, 4).

Now, let us find the slope of the given line,

$\text { Slope of a line }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{1}=-2$

$\text { Slope of a line } \times \text { slope of perpendicular line }=-1$

$\begin{array}{l}{-2 \times \text { slope of perpendicular line }=-1} \\\\ {\text { Slope of perpendicular line }=-1 \times \frac{1}{-2}=\frac{1}{2}}\end{array}$

Now, slope of our required line = 1/2 and it passes through (5, 4)

The point slope form is given as:

$y-y_{1}=m\left(x-x_{1}\right)$

$\text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on the line. }$

$\text { Here in our problem, } \mathrm{m}=\frac{1}{2}, \text { and }\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(5,4)$

$y-4=\frac{1}{2}(x-5)$

Now let us convert to standard form:

The standard form of a line is just another way of writing the equation of a line.

The standard form of an equation is Ax + By = C. In this kind of equation, x and y are variables and A, B, and C are integers.

2(y – 4) = 1(x - 5)

2y – 8 = x – 5

x – 2y - 5 + 8 = 0

x - 2y + 3 = 0

x - 2y = - 3

Hence, the line equation in standard form is x - 2y = - 3

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3 years ago

The College Student Journal (December 1992) investigated differences in traditional and nontraditional students, where nontradit

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Answer:

94.52% probability that the random sample of 100 nontraditional students has a mean GPA greater than 3.42.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3.5, \sigma = 0.5, n = 100, s = \frac{0.5}{\sqrt{100}} = 0.05$

What is the probability that the random sample of 100 nontraditional students has a mean GPA greater than 3.42?

This is 1 subtracted by the pvalue of Z when X = 3.42.

$Z = \frac{X - \mu}{s}$

$Z = \frac{3.42 - 3.5}{0.05}$

$Z = -1.6$

$Z = -1.6$ has a pvalue of 0.0548.

So there is a 1-0.0548 = 0.9452 = 94.52% probability that the random sample of 100 nontraditional students has a mean GPA greater than 3.42.

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4 years ago