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Kipish [7]
3 years ago
15

(03.06) Given the equation y − 3 = one half (x + 6) in point-slope form, identify the equation of the same line in slope interce

pt form. (1 points) x − 2y = −12 y = one halfx + 6 y = one halfx + 9 y = one halfx
Mathematics
1 answer:
Anvisha [2.4K]3 years ago
5 0
Y-3=(1/2)(x+6). just solve for y
y-3=1/2x +3
y = 1/2x + 6

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2 years ago
Pam draws three scalene triangles. In each figure, she measures each angle, as shown.
Ganezh [65]

Answer:

the answer is c.

Step-by-step explanation:

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7 0
3 years ago
The area of a circle is 16π cm2. What is the circle's circumference?
agasfer [191]

Answer:

C = 8 pi cm

Step-by-step explanation:

The area of a circle is given by

A = pi r^2

16 pi = pi r^2

Divide each side by pi

16 = r^2

Taking the square root

4 = r

The circumference is

C = 2 * pi *r

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C = 8 pi cm

4 0
2 years ago
Use the graph below to estimate the solution to the system of equations shown.
Mila [183]

Answer: The second choice is answer.

Step-by-step explanation:

<h3>The first line (Blue One)</h3>

y=-x-3 (Slope is -1 and intercepts y at -3)

<h3>The second one (Red One)</h3>

y=2x-8 (Slope is 2, to find y-intercept. Substitute x = 0 then we get -8) (For x-intercept, it's 4. Just substitute y=0 to get x-intercept.)

Because there are choices, it's easy to notice.

The first one, x = -4 and almost -5 which is not right. (x is around 1 almost 2.)

The second one, x = 1 and almost 2 is right.

The third one, x = -1 and almost -2 is not right (The intersection isn't even at -1 or any negative number for x-axis.)

The fourth one, just like the first one... x = 4 almost 5

<h3>Or solve the equations.</h3>

y=-x-3\\y=2x-8

Substitute y=2x-8 in y=-x-3

2x-8=-x-3\\2x-8+x+3=0\\3x-5=0\\3x=5\\x=\frac{5}{3}\\(x=1\frac{2}{3})

Look at the value of x then find the answer that matches the value of x.

Now for y, substitute x = 5/3 in y = -x-3 (or in 2x-8 if you want.)

y=-\frac{5}{3} -3\\y=-\frac{5}{3} -\frac{9}{3} \\y=-\frac{14}{3} \\(y=-4\frac{2}{3} )

So the answer is the second choice.

6 0
3 years ago
2. The time between engine failures for a 2-1/2-ton truck used by the military is
OLEGan [10]

Answer:

A truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.

For a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

Step-by-step explanation:

We have here a <em>random variable</em> <em>normally distributed</em> (the time between engine failures). According to this, most values are around the mean of the distribution and less are far from it considering both extremes of the distribution.

The <em>normal distribution</em> is defined by two parameters: the population mean and the population standard deviation, and we have each of them:

\\ \mu = 6000 miles.

\\ \sigma = 800 miles.

To find the probabilities asked in the question, we need to follow the next concepts and steps:

  1. We will use the concept of the <em>standard normal distribution</em>, which has a mean = 0, and a standard deviation = 1. Why? With this distribution, we can easily find the probabilities of any normally distributed data, after obtaining the corresponding <em>z-score</em>.
  2. A z-score is a kind of <em>standardized value</em> which tells us the <em>distance of a raw score from the mean in standard deviation units</em>. The formula for it is: \\ z = \frac{x - \mu}{\sigma}. Where <em>x</em> is the value for the raw score (in this case x = 5000 miles).
  3. The values for probabilities for the standard normal distribution are tabulated in the <em>standard normal table</em> (available in Statistics books and on the Internet). We will use the <em>cumulative standard normal table</em> (see below).

With this information, we can solve the first part of the question.

The chance that a truck will be able to travel a total distance of over 5000 miles without an engine failure

We can "translate" the former mathematically as:

\\ P(x>5000) miles.

The z-score for x = 5000 miles is:

\\ z = \frac{5000 - 6000}{800}

\\ z = \frac{-1000}{800}

\\ z = -1.25

This value of z is negative, and it tells us that the raw score is 1.25 standard deviations <em>below</em> the population mean. Most standard normal tables are made using positive values for z. However, since the normal distribution is symmetrical, we can use the following formula to overcome this:

\\ P(z

So

\\ P(z

Consulting a standard normal table available on the Internet, we have

\\ P(z

Then

\\ P(z1.25)

\\ P(z1.25)

However, this value is for P(z<-1.25), and we need to find the probability P(z>-1.25) = P(x>5000) (Remember that we standardized x to z, but the probabilities are the same).

In this way, we have

\\ P(z>-1.25) = 1 - P(z

That is, the complement of P(z<-1.25) is P(z>-1.25) = P(x>5000). Thus:

\\ P(z>-1.25) = 1 - 0.10565

\\ P(z>-1.25) = 0.89435  

In words, a truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.

We can see the former probability in the graph below.  

The chance that a fleet of a dozen trucks will have an average time-between-failures of 5000 miles or more

We are asked here for a sample of <em>12 trucks</em>, and this is a problem of <em>the sampling distribution of the means</em>.

In this case, we have samples from a <em>normally distributed data</em>, then, the sample means are also normally distributed. Mathematically:

\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})

In words, the samples means are normally distributed with the same mean of the population mean \\ \mu, but with a standard deviation \\ \frac{\sigma}{\sqrt{n}}.

We have also a standardized variable that follows a standard normal distribution (mean = 0, standard deviation = 1), and we use it to find the probability in question. That is

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z \sim N(0, 1)

Then

The "average time-between-failures of 5000" is \\ \overline{x} = 5000. In other words, this is the mean of the sample of the 12 trucks.

Thus

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z = \frac{5000 - 6000}{\frac{800}{\sqrt{12}}}

\\ z = \frac{-1000}{\frac{800}{\sqrt{12}}}

\\ z = \frac{-1000}{230.940148}

\\ z = -4.330126

This value is so low for z, that it tells us that P(z>-4.33) is almost 1, in other words it is almost certain that for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is almost 1.

\\ P(z

\\ P(z

\\ P(z

The complement of P(z<-4.33) is:

\\ P(z>-4.33) = 1 - P(z or practically 1.

In conclusion, for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

7 0
3 years ago
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