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2 years ago

The ages of 8 different students in a college math class were used to make the box plot shown above. Which of the following

Mathematics

1 answer:

notsponge [240]2 years ago

8 0

Answer:

Range is 8

Step-by-step explanation:

To find range you need to subtract the largest number with the smallest. The largest is 28 and the smallest is 20.

28-20=8

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Solve the following system of equations.<br><br> 2x+y=3<br> x=2y-1

Anvisha [2.4K]

Answer:

Step-by-step explanation:

1,1

5 0

2 years ago

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I would like to know what is the value of z

Brrunno [24]

From the chord theorem we know that:

$\overline{TW}\cdot\overline{UW}=\overline{CW}\cdot\overline{VW}$

so:

$\overline{TW}\cdot\overline{UW}=\overline{CW}\cdot\overline{VW}\\\\3\cdot\overline{UW}=x\cdot6\\\\3\cdot(\overline{TU}-\overline{TW})=x\cdot6\\\\
3\cdot(x+7-3)=x\cdot6\\\\3\cdot(x+4)=x\cdot6\\\\3x+12=6x\\\\12=6x-3x\\\\3x=12\qquad|\div3\\\\\boxed{x=4}$

Answer C.

3 0

3 years ago

Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6

san4es73 [151]

Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

So, we have:

$f(x)= \frac{9}{3x+ 2}$

Rewrite as:

$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

Factorize

$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

2 years ago

Which value is equivalent to 48 ÷ ((2 + 6) × 2) − 1? A) -12 B) 2 C) 3 1 5 D) 3

timama [110]

Answer:

B) 2

Step-by-step explanation:

Given expression,

48 ÷ ((2 + 6) × 2) − 1

In simplification we follows the following order,

B = Bracket

O = Of

D = Division

M = Multiplication

A = Addition,

S = Subtraction,

Thus, steps of solving the given expression by following BODMAS are as follow,

$\frac{48}{(2 + 6) \times 2}-1$

$=\frac{48}{8\times 2}-1$

$=\frac{48}{16}-1$

$=3-1$

$=2$

Hence, OPTION B is correct.

6 0

3 years ago

PLZ HELP!! 50 POINTS!

yKpoI14uk [10]

x = 3

Step-by-step explanation:

5x = 3x + 6

5x - 3x = 6

2x = 6

x = 6/2

x = 3

6 0

3 years ago

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