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3 years ago

Translate the graph of the equation y=5/2 x+2 to the left 4 units. Graph the translation

Mathematics

1 answer:

ahrayia [7]3 years ago

8 0

Answer:

$\large\boxed{y=\dfrac{5}{2}x+12}$

Step-by-step explanation:

n > 0

f(x + n) - shift the graph n units to the left

f(x - n) - shift the graph n units to the right

f(x) + n - shift the graph n units up

f(x) - n - shift the graph n units down

$y=\dfrac{5}{2}x+2\to f(x)=\dfrac{5}{2}x+2$

translate the graph 4 units to the left:

$f(x+4)=\dfrac{5}{2}(x+4)+2$ use the distributive property

$f(x+4)=\left(\dfrac{5}{2}\right)(x)+\left(\dfrac{5}{2}\right)(4)+2\\\\f(x+4)=\dfrac{5}{2}x+(5)(2)+2\\\\f(x+4)=\dfrac{5}{2}x+10+2\\\\f(x+4)=\dfrac{5}{2}x+12$

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3 years ago

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3 years ago

A jar contains 11red marbles, 12 blue marbles and 6 white marbles. Four marbles from the jar are selected. With each marble bein

tangare [24]

Answer:

Red on the 5th draw = 0.0907

Step-by-step explanation:

The first to fourth selections are all the same.

Blue + white = 12 + 6 = 18

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3 years ago

What are the solutions to the nonlinear system of equations below?

Sonbull [250]

Substitute y=4x to the second equation:

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x^2 + 16x^2 = 17
17x^2 = 17
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When x=1, y=4(1) = 4
When x=-1, y=4(-1) = -4

Thus the solutions would be (1,4) and (-1,-4). That would correspond to D. and A.

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3 years ago

By law, an industrial plant can discharge not more than 500 gallons of waste water per hour, on the average, into a neighboring

Alisiya [41]

Answer:

We conclude that not more than 500 gallons of wastewater per hour, on average, is discharged into a neighboring lake.

Step-by-step explanation:

We are given that an industrial plant can discharge not more than 500 gallons of wastewater per hour, on average, into a neighboring lake.

Four one-hour periods are selected randomly over a period of one week. The following are observed:

1384, 683, 1534, 405

Let $\mu$ = population average gallons of wastewater discharged per hour

So, Null Hypothesis, $H_0$ : $\mu \leq$ 500 gallons {means that not more than 500 gallons of wastewater per hour, on the average, is discharged into a neighboring lake}

Alternate Hypothesis, $H_A$ : $\mu$ > 500 gallons {means that more than 500 gallons of wastewater per hour, on the average, is discharged into a neighboring lake}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 1001.5 gallons

s = sample standard deviation = $\sqrt{\frac{\sum (X - \bar X)^{2} }{n-1} }$ = 543.79

n = sample of periods = 4

So, the test statistics = $\frac{1001.5-500}{\frac{543.79}{\sqrt{4} } }$ ~ $t_3$

= 1.844

The value of t-test statistics is 1.844.

Since in the question we are not given with the level of significance so we assume it to be 5%. Now, at 0.05 level of significance, the t table gives a critical value of 2.353 at 3 degrees of freedom for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 1.844 < 2.353, so we have insufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that not more than 500 gallons of wastewater per hour, on average, is discharged into a neighboring lake.

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3 years ago