What is the multiplicative inverse of 2/3?

A circle is centered at the point (5,-4) & passes through the point (-3,2).

maxonik [38]

Equation of the circle:

$(x-5)^2+(y+4)^2=100$

Step-by-step explanation:

The equation of a circle is given by

$(x-x_0)^2+(y-y_0)^2=r^2$

where

$(x_0,y_0)$ are the coordinates of the center of the circle

r is the radius of the circle

For the circle in this problem, we know that:

(5,4) are the coordinates of the center

We also know that the circle passes through the point (-3,2), therefore we can calculate the radius as the distance between the point on the circle and the center of the circle:

$r=\sqrt{(-3-(5))^2+(2-(-4))^2}=\sqrt{(-8)^2+(6)^2}=\sqrt{64+36}=10$

Therefore, the equation of the circle is:

$(x-5)^2+(y-(-4))^2=10^2\\(x-5)^2+(y+4)^2=100$

Learn more about circles:

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#LearnwithBrainly

5 0

3 years ago

Y=74(1.01)^8 is it increasing or decreasing

Dvinal [7]

Answer:

Increasing

Step-by-step explanation:

The computation is given below:

Given that

y = 74 × (1.01)^8

= 74 × 1.082856706

= 80.1314

As it can be seen that the initial amount is 74 but after solving the given equation the value is increased as it shows 80.1314

Therefore it is increasing

5 0

3 years ago

$1725 with 15% markup what’s the new price

fredd [130]

Answer:

$1983.75

Step-by-step explanation:

1725 x 0.15 = 258.75

now add to the original price

1725 + 258.75= 1983.75

5 0

3 years ago

Give an example of an absolute value inequality whose answers falls between two values. Give an example of an absolute value ine

mel-nik [20]

Answer:

A) Suppose we have an onedimensional situation.

in the 0 of our x-axis, we have a fruit tree, and we want to rest at a distance no bigger than 6 ft of the tree, then all the possible positions of our resting place are:

x ∈(-6ft, 6ft)

we can write this as: IxI < 6ft

b) now we think the opposite situation, we want to rest at least 6ft away from the tree, then we have that:

x ∉ [-6ft, 6ft].

or IxI > 6ft.

So you can see that the difference in those two cases is if we want to be "inside a given range" (for the first case) or "outside a given range" (for the second case).

8 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 "will be able to travel a total distance of over 5000 miles without an engine failure" 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 random variable normally distributed (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 normal distribution 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:

We will use the concept of the standard normal distribution, 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 z-score.
A z-score is a kind of standardized value which tells us the distance of a raw score from the mean in standard deviation units. The formula for it is: $\\ z = \frac{x - \mu}{\sigma}$ . Where x is the value for the raw score (in this case x = 5000 miles).
The values for probabilities for the standard normal distribution are tabulated in the standard normal table (available in Statistics books and on the Internet). We will use the cumulative standard normal table (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 below 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

$\\ P(z1.25)$

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 "will be able to travel a total distance of over 5000 miles without an engine failure" 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 12 trucks, and this is a problem of the sampling distribution of the means.

In this case, we have samples from a normally distributed data, 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.

$\\ P(z$

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