All categories

3 years ago

Does the point (10, 13) satisfy the equation y = x?

Mathematics

2 answers:

Trava [24]3 years ago

8 0

No, (10,13) does not satisfy the equation y=x

Hope this helps

Trava [24]3 years ago

7 0

Equation: y = x

When x = 10, y = 10

----------------------------------------------------------------------------
Answer: No, (10, 13) does not satisfy the equation y = x.
---------------------------------------------------------------------------

You might be interested in

If we sample from a small finite population without replacement, the binomial distribution should not be used because the event

seropon [69]

Answer:

5/4324 = 0.001156337

Step-by-step explanation:

To better understand the hyper-geometric distribution consider the following example:

There are 100 senators in the US Congress, and suppose 60 of them are republicans so 100 - 60 = 40 are democrats).

We extract a random sample of 30 senators and we want to answer this question:

What is the probability that 10 senators in the sample are republicans (and of course, 30 - 10 = 20 democrats)?

The answer using the h-g distribution is:

$\large \frac{\binom{60}{10}\binom{100-60}{30-10}}{\binom{100}{30}}=\frac{\binom{60}{10}\binom{40}{20}}{\binom{100}{30}}$

Now, imagine there are 56 senators (56 lottery numbers), 6 are republicans (6 winning numbers and 50 losers), we extract a sample of 6 senators (the bettor selects 6 numbers). What is the probability that 4 senators are republicans? (What is the probability that 4 numbers are winners?).

<em>As we see, the situation is exactly the same,</em> but changing the numbers. So the answer would be

$\large \frac{\binom{6}{4}\binom{56-6}{6-4}}{\binom{56}{6}}=\frac{\binom{6}{4}\binom{50}{2}}{\binom{56}{6}}$

Now compute each combination separately:

$\large \binom{6}{4}=\frac{6!}{4!2!}=15\\\\\binom{50}{2}=\frac{50!}{2!48!}=1225\\\\\binom{50}{6}=\frac{50!}{6!44!}=15890700$

and now replace the values:

$\large \frac{\binom{6}{4}\binom{50}{2}}{\binom{56}{6}}=\frac{15*1225}{15890700}=\frac{18375}{15890700}=\frac{5}{4324}$

and that is it.

If the decimal expression is preferred then divide the fractions to get 0.001156337

6 0

3 years ago

384,000 in scientific notaion

bixtya [17]

Answer: 3.84 · 10^5

Step-by-step explanation:

8 0

3 years ago

Explain why any of the four operations placed between two terms 5 and -3 sqrt (8) will result in an irrational number

Gnoma [55]

Sum/difference:

Let

$x = 5 + (-3\sqrt{8}) = 5-3\sqrt{8}$

This means that

$3\sqrt{8} = 5-x \iff \sqrt{8} = \dfrac{5-x}{3}$

Now, assume that $x$ is rational. The sum/difference of two rational numbers is still rational (so 5-x is rational), and the division by 3 doesn't change this. So, you have that the square root of 8 equals a rational number, which is false. The mistake must have been supposing that $x$ was rational, which proves that the sum/difference of the two given terms was irrational

Multiplication/division:

The logic is actually the same: if we multiply the two terms we get

$x = -15\sqrt{8}$

if again we assume x to be rational, we have

$\sqrt{8} = -\dfrac{x}{15}$

But if x is rational, so is -x/15, and again we come to a contradiction: we have the square root of 8 on one side, which is irrational, and -x/15 on the other, which is rational. So, again, x must have been irrational. You can prove the same claim for the division in a totally similar fashion.

7 0

2 years ago

ALGEBRAIC EXPRESSION 11. Subtract the sum of 13x – 4y + 7z and – 6z + 6x + 3y from the sum of 6x – 4y – 4z and 2x + 4y – 7. 12.

Naily [24]

Answer:

Explained below.

Step-by-step explanation:

(11)

Subtract the sum of (13x - 4y + 7z) and (- 6z + 6x + 3y) from the sum of (6x - 4y - 4z) and (2x + 4y - 7z).

$[(6x - 4y - 4z) +(2x + 4y - 7z)]-[(13x - 4y + 7z) + (- 6z + 6x + 3y) ]\\=[6x-4y-4z+2x+4y-7z]-[13x-4y+7z-6z+6x+3y]\\=6x-4y-4z+2x+4y-7z-13x+4y-7z+6z-6x-3y\\=(6x+2x-13x-6x)+(4y-4y+4y-3y)-(4z+7z+7z-6z)\\=-11x+y-12z$

Thus, the final expression is (-11x + y - 12z).

(12)

From the sum of (x² + 3y² - 6xy), (2x² - y² + 8xy), (y² + 8) and (x² - 3xy) subtract (-3x² + 4y² - xy + x - y + 3).

$[(x^{2} + 3y^{2} - 6xy)+(2x^{2} - y^{2} + 8xy)+(y^{2} + 8)+(x^{2} - 3xy)] - [-3x^{2} + 4y^{2} - xy + x - y + 3]\\=[x^{2} + 3y^{2} - 6xy+2x^{2} - y^{2} + 8xy+y^{2} + 8+x^{2} - 3xy]- [-3x^{2} + 4y^{2} - xy + x - y + 3]\\=[4x^{2}+3y^{2}-xy+8]-[-3x^{2} + 4y^{2} - xy + x - y + 3]\\=4x^{2}+3y^{2}-xy+8+3x^{2}-4y^{2}+xy-x+y-3\\=7x^{2}-y^{2}-x+y+5$