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3 years ago

Find an equation of the line passing through each of the following pairs of points. a (−3, 1), (0, 3)

1 answer:

5 0

$\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-3)}}}\implies \cfrac{2}{0+3}\implies \cfrac{2}{3}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{\cfrac{2}{3}}[x-\stackrel{x_1}{(-3)}]\implies y-1=\cfrac{2}{3}(x+3) \\\\\\ y-1=\cfrac{2}{3}x+2\implies y=\cfrac{2}{3}x+3$

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Complete the following sentence. A radius is________ the diameter.

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Answer:

1/2

Step-by-step explanation:

The radius is 1/2 of the diameter

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3 years ago

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Determine the range of the function.

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Answer:

$Solution: f(x)\ge-5\\Interval:[-5,\infty)$

Step-by-step explanation:

Equation: $\frac{1}{4} x^2-5 = y$

1). $\frac{1}{4} x^2-5 = y$ → $\frac{x^2}{4}-5=y$

2). $\frac{x^2}{4}-5=y$ ∴ $a=\frac{1}{4}, b=0, c=-5$

3). $x_v=-\frac{b}{2a}$

$=-\frac{0}{2(\frac{1}{4})}$

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$y_v=\frac{0^2}{4}-5$

$=-5$

5). Minimum (0,-5) ∴ $f(x)\ge-5$

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3 years ago

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3 years ago

A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer

pshichka [43]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.

The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.

A sample of 34 cups was taken:

a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:

The lower value will be:

$Z_{o.o5}$ = -1.648

You reverse the standardization using the formula $Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)$

$-1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }$

X[bar]= 7.72ounces

The lower control point will be 7.72 ounces.

The upper value will be:

$Z_{0.95}= 1.648$

$1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }$

X[bar]= 8.28ounces

The upper control point will be 8.82 ounces.

b. Now μ= 7.6, considering the control limits of a.

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-7.6)/(1/√34))- P(Z≤7.72-7.6)/(1/√34))

P(Z≤7.11)- P(Z≤0.70)= 1 - 0.758= 0.242

There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.

b. For this item μ= 8.7, the control limits do not change:

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))

P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007

There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.

I hope it helps!

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3 years ago

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Consider the following equation: f(x)=x^2+4\4x^2-4x-8 name the vertical asymptote(s)

zhenek [66]

ANSWER