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ioda
3 years ago
11

What is the question of the line that passes through the point (-2,-2) and has a slope of 4. Plz someone answer

Mathematics
1 answer:
Studentka2010 [4]3 years ago
4 0

Answer:y=4x+6

Step-by-step explanation:

We have the information that the slope is 4 and the line goes through the point (-2,-2). With this information, we can make a linear equation in a point slope form (y - y_{1}= m * (x - x_{1}), so the equation would be y+2=4(x+2), or simplified, y+2=4x+8. in order to solve for y (to make it a slope-intercept equation), we must subtract 2 from both sides. This gives us the equation y=4x+6. Hope this helps!

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babunello [35]
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2 years ago
Assume that a company sold 5.75 million motorcycles and 3.5 million cars in the year 2010. The growth in the sale of motorcycles
OverLord2011 [107]

Answer:

Final answer is approx 6.644 years.

Step-by-step explanation:

Given that a company sold 5.75 million motorcycles and 3.5 million cars in the year 2010. The growth in the sale of motorcycles is 16% every year and that of cars is 25% every year.

So we can use growth formula:

A=P\left(1+r\right)^t

Then we get equation for motorcycles and cars as:

A=5.75\left(1+0.16\right)^t

A=3.5\left(1+0.25\right)^t

Now we need to find about when the sale of cars will be more than the sale of motorcycles. So we get:

3.5\left(1+0.25\right)^t>5.75\left(1+0.16\right)^t

3.5\left(1.25\right)^t>5.75\left(1.16\right)^t

3.5\left(1.25\right)^t>5.75\left(1.16\right)^t

\frac{\left(1.25\right)^t}{\left(1.16\right)^t}>\frac{5.75}{3.5}

\left(\frac{1.25}{1.16}\right)^t>1.64285714286

t\cdot\ln\left(\frac{1.25}{1.16}\right)>\ln\left(1.64285714286\right)

t>\frac{\ln\left(1.64285714286\right)}{\ln\left(\frac{1.25}{1.16}\right)}

t>6.6436473051

Hence final answer is approx 6.644 years.

7 0
2 years ago
What is the area of the polygon below
kvasek [131]
The area is 84.

What I do it cut the polygon into something easier like a square and rectangle.

Rectangle: 6•12=72
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Add them together and you have 84
5 0
2 years ago
The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for
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f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}

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f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982

The probability of doing so for at least 2 of 5 days is

\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x

Then the desired probability is

F_X(30)-F_X(15)\approx0.950-0.777=0.173

7 0
3 years ago
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Do you have the answer options? Sorry, can't answer without them ):
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