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3 years ago

10

The ace telephone co. charges a flat monthly fee of $22.00 for a telephone line and $0.20 per minute for long distance calls. Sk

etch a graph of this equation for up to 100 minutes of calls

1 answer:

Trava [24]3 years ago

7 0

$y = .2x + 22$

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I need help with this problem

Monica [59]

B)7:8 is the answer because:

16 total - 7 green = 8 and so
the ratio of GREEN to RED is
7:8
7 green 8 red

For a) it would be 1:2 as it is 16:8 and simplified would make it 1:2

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3 years ago

Select the true statement for the graph

salantis [7]

Answer:

C srry if wrong

Step-by-step explanation:

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2 years ago

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Simplify (2¼-1⅔)×1⅝ please. The ⅝ the 8is 7

bekas [8.4K]

Answer:

I don't understand the part "the 5/8 the 8is7" ?????????????

Step-by-step explanation:

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2 years ago

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A new method for timing traffic signals in heavily traveled intersections is introduced. The effectiveness of the new method was

Charra [1.4K]

Answer:

The sample size must be approximately 211.

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = 654.1

Alpha, α = 0.02

Population standard deviation, σ = 311.7

Margin of error = 50

Formula:

$z_{critical}\frac{\sigma}{\sqrt{n}}$

$z_{critical}\text{ at}~\alpha_{0.02} = \pm 2.33$

Putting all the values, we get,

$\pm 50 = \pm 2.33 (\dfrac{311.7}{\sqrt{n}} ) \\\\\sqrt{n} = \dfrac{2.33\times 311.7}{50}\\\\n = 210.98\\n \approx 211$

Thus, the sample size must be approximately 211.

6 0

3 years ago

Use the denition of the derivative to find f 0 (3), where f (x) = 3x+5 / 2x−1

Crazy boy [7]

Answer:

$f'(3)= -\frac{13}{25}$

Step-by-step explanation:

We are asked to find $f'(3)$ of function $f(x)=\frac{3x+5}{2x-1}$ using definition of derivatives.

Limit definition of derivatives:

$f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Let us find $f(3+h)$ and $f(3)$ .

$f(3+h)=\frac{3(3+h)+5}{2(3+h)-1}$

$f(3+h)=\frac{9+3h+5}{6+2h-1}\\\\f(3+h)=\frac{3h+14}{2h+5}$

$f(3)=\frac{3(3)+5}{2(3)-1}$

$f(3)=\frac{9+5}{6-1}\\\\f(3)=\frac{14}{5}$

Substituting these values in limit definition of derivatives, we will get:

$f'(3)= \lim_{h \to 0} \frac{f(3+h)-f(3)}{h}$

$f'(3)= \lim_{h \to 0} \frac{\frac{3h+14}{2h+5}-\frac{14}{5}}{h}$

Make a common denominator:

$f'(3)= \lim_{h \to 0} \frac{\frac{(3h+14)*5}{(2h+5)*5}-\frac{14(2h+5)}{5(2h+5)}}{h}$

$f'(3)= \lim_{h \to 0} \frac{\frac{5(3h+14)-14(2h+5)}{5(2h+5)}}{h}$

$f'(3)= \lim_{h \to 0} \frac{5(3h+14)-14(2h+5)}{5h(2h+5)}$

$f'(3)= \lim_{h \to 0} \frac{15h+70-28h-70}{5h(2h+5)}$

$f'(3)= \lim_{h \to 0} \frac{-13h}{5h(2h+5)}$

Cancel out h:

$f'(3)= \lim_{h \to 0} \frac{-13}{5(2h+5)}$

$f'(3)= \frac{-13}{5(2(0)+5)}$

$f'(3)= \frac{-13}{5(5)}$

$f'(3)= -\frac{13}{25}$

Therefore, $f'(3)= -\frac{13}{25}$ .

8 0

3 years ago