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3 years ago

The points (-6, r) and (6,3) lie on a line with slope 3/4. Find the missing coordinate r.

Mathematics

1 answer:

NeTakaya3 years ago

5 0

Answer:

r = -6

Step-by-step explanation:

Equation of a line:

The equation of a line has the following format:

$y = mx + b$

In which m is the slope and b is the y-intercept.

Slope 3/4

This means that $m = \frac{3}{4}$ , and thus:

$y = \frac{3}{4}x + b$

(6,3)

This means that when $x = 6, y = 3$ . We use this to find b. So

$y = \frac{3}{4}x + b$

$3 = \frac{3}{4}(6) + b$

$b = 3 - \frac{18}{4} = \frac{12}{4} - \frac{18}{4} = -\frac{6}{4}$

$y = \frac{3}{4}x - \frac{6}{4} = \frac{3x - 6}{4}$

(-6, r)

r is y when $x = -6$ . So

$y = \frac{3(-6) - 6}{4} = -6$

So r = -6.

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Answer: k = 6

Step-by-step explanation: Isolate the variable by dividing each side by factors that don't contain the variable.

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3 years ago

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2 years ago

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

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3 years ago

If rectangle ABCD was reflected over the y-axis, reflected over the x-

olga_2 [115]

Answer:

-4, 1

Step-by-step explanation:

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3 years ago

At an ocean depth of 8 meters, a buoy bobs up and then down 5 meters from the ocean's depth. Sixteen seconds pass from the time

lisov135 [29]

Answer:

See attached for a graph

Step-by-step explanation:

We're going to plot sea level as y=0 and a depth of 8 meters as y=-8.

The problem statement tells you the initial point (x=0) is at normal ocean depth (y=-8), so the first point you put into your sine tool is ...

(x, y) = (0, -8)

The buoy takes 16 seconds to go from a high point to a low point, so the time to the first high point is half that, or x=8 seconds. That high point is 5 meters above its average depth, so is at y=-3.

The second point you will put into your sine tool is ...

(x, y) = (8, -3)

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3 years ago