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2 years ago

Select the real-world problem that could be solved using a proportion.

Mathematics

1 answer:

AlekseyPX2 years ago

5 0

Answer:

Step-by-step explanation:

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Use the Counting Principle to find the probability. Choosing the 7 winning lottery numbers when the numbers are chosen at random

frozen [14]

Answer: 120

Step-by-step explanation:

Since the order of the numbers doesn't matter we can use the formula:

$\dfrac{n!}{r!(n-r)!}\quad \text{where n is the quantity of items and r is the quantity chosen}\\\\\text{In this problem, there are 10 numbers (n = 10) and 7 to be chosen (r = 7)}\\\\_{10}C_{7}=\dfrac{10!}{7!(10-7)!}\\\\.\qquad=\dfrac{10!}{7!3!}\\\\.\qquad=\dfrac{10\times 9\times 8\times 7!}{7!\times 3\times 2\times 1}\\\\.\qquad=10\times 3\times 4\\\\.\qquad=120$

7 0

3 years ago

Read 2 more answers

Can anyone please help me with this???

ira [324]

Answer:

x=8a

Step-by-step explanation:

(3/a) x - 4 = 20

add 4 to both sides

(3/a) x = 24

Divide by 3/a

x = 24 / (3/a)

Use KCF (keep change flip)

x = (24/1) x (a/3)

x = 24a / 3

simplify

x = 8a

3 0

3 years ago

“ write all your answers with a positive exponent so your answer will be in the form of x^# or 1/(x^#)

gayaneshka [121]

Answer:

Step-by-step explanation:

5). $\frac{1}{x^{-9}}$ = $\frac{1}{\frac{1}{x^9} }$

= $x^{9}$

6). $\frac{1}{x^{-2}}=\frac{1}{\frac{1}{x^2} }$

= x²

7). $x^{-6}=\frac{1}{x^6}$

8). $\frac{1}{x^{-7}}=\frac{1}{\frac{1}{x^7} }$

= $x^{7}$

9). $x^{-8}=\frac{1}{x^8}$

10). $\frac{1}{x^{-3}}$ = $\frac{1}{\frac{1}{x^{3}}}$

= x³

7 0

3 years ago

Find the center of the circle that you can circumscribe about DABC.

Vlada [557]

The center of the circle that circumscribe about DABC is $(h,k) = (1, 5)$ .

A circle can be modelled after a expression of the form:

$A\cdot x + B\cdot y + C =-x^{2}-y^{2}$ (1)

We can determine all coefficients by knowing three distinct points on plane.

If we know that $(x_{1}, y_{1}) = (2, 8)$ , $(x_{2}, y_{2}) = (0, 8)$ and $(x_{3}, y_{3}) = (2, 2)$ , then the solution of the system of linear equations is:

$2\cdot A + 8\cdot B + C = -68$ (2)

$8\cdot B + C = -64$ (3)

$2\cdot A + 2\cdot B + C = -8$ (4)

$A = -2, B = -10, C = 16$

Now we proceed to complete squares and factor each resulting perfect square trinomial in order to determine the coordinates of the center of the circle:

$x^{2}+y^{2}-2\cdot x -10\cdot y +16 = 0$