Ask question

Ask question

All categories

3 years ago

6

The measures of the interior angles of a pentagon are 2x,6x,4x-6,2x-16 and 6x+2. What is the measure, in degrees, of the largest

angle?

1 answer:

Varvara68 [4.7K]3 years ago

7 0

Answer:

170°

Step-by-step explanation:

The formulae for getting the interior sides of an angle is 180(n-2) where n is there number of sides. And pentagon has 5 sides.

180(5-2)

180(3)=540°

Additions of the must just give 540°

2x+6x+(4x-6)+(2x-16)+(6x+2)=540°

Collect like terms

2x+6x+4x+2x+6x-6-16+2=540°

20x-20=540

20x=540+20

20x=560

x=560/20

x=28°

Which of these sides gives the greatest.

2x=2(28)=56°

6x=6(28)=168°

4x-6=4(28)-6=112-6=108°

2x-16=2(28)-16=56-15=40°

6x+2=6(28)+2=168+2=170°

The greatest of them is (6x+2)°=170°

You might be interested in

Expand and simplify (x-6)(x+2)

maw [93]

I hope this helps you

4 0

3 years ago

Read 2 more answers

What is the area of 2 yards

enot [183]

The area of two yards is 18ft

8 0

3 years ago

at Phil's work there are 12 part time workers and 18 full time workers what is the ratio of part time workers to total workers

BabaBlast [244]

Answer:

the ratio of part time workers to total workers are 12/30

3 0

3 years ago

"A manufacturer of automobile batteries claims that the average length of life for its grade A battery is 55 months. Suppose the

STALIN [3.7K]

Answer:

$\\ P(z>-2) = 0.97725$ or P(x>49) is about 97.725% (or being less precise 97.5% using the <em>empirical rule</em>).

Step-by-step explanation:

We solve this question using the following information:

We are dealing here with <em>normally distributed data</em>, that is "<em>the frequency distribution of the life length data is known to be mound-shaped</em>".
The normal distribution is defined by two parameters: the population mean ( $\\ \mu$ ) and the population standard deviation ( $\\ \sigma$ ). In this case, we have that $\\ \mu = 55$ months, and $\\ \sigma = 3$ months.
To find the probabilities, we have to use the <em>standard normal distribution</em>, which has $\\ \mu = 0$ and $\\ \sigma = 1$ . The probabilities for this distribution are collected in the <em>standard normal table</em>, available in Statistics books or on the Internet. We can also use statistics programs to find these probabilities.
For most cases, we need to use the <em>cumulative standard normal table, </em>and for this we have to previously "transform" a raw score (x) into a z-score using the next formula: $\\ z = \frac{x - \mu}{\sigma}$ [1]. A z-score tells us the distance from the mean that a raw score is from it in <em>standard deviations units</em>. If this value is <em>negative</em>, the raw score is <em>below</em> the mean. Conversely, a <em>positive</em> value indicates that it is <em>above</em> the mean.
The <em>cumulative standard normal table </em>is made for positive values of z. Since the normal distribution is <em>symmetrical</em> around the mean, we can find the negative values of z using this formula: $\\ P(z$ [2].

Having all this information, we can solve the question.

<h3>The percentage of the manufacturer's grade A batteries that will last more than 49 months</h3>

<em>First Step: Use formula [1] to find the z-score of the raw score x = 49 months</em>.

$\\ z = \frac{49 - 55}{3}$

$\\ z = \frac{-6}{3}$

$\\ z = -2$

This means that the raw score is represented by a z-score of $\\ z = -2$ , which tells us that it is<em> two standard deviations below</em> the population mean.

<em>Second Step: Consult this value in the cumulative standard normal table for z = 2 and apply the formula [2] to find the corresponding probability.</em>

For a z = 2, the probability is 0.97725.

Then

$\\ P(z$

$\\ P(z2)$

$\\ P(z2)$

But we <em>are not asked</em> for P(z<-2) but for P(z>-2) = P(x>49). This probability is the <em>complement</em> of the previous result, that is

$\\ P(z>-2) = 1 - P(z$

$\\ P(z>-2) = 1 - 0.02275$

$\\ P(z>-2) = 0.97725$

That is, the "<em>percentage of the manufacturer's grade A batteries will last more than 49 months</em>" is

$\\ P(z>-2) = 0.97725$ or about 97.725%

A graph below shows this result.

Notice that if we had used the <em>68-95-99.7 rule</em> (also known as the <em>empirical rule</em>), that is, in a normal distribution, the interval between <em>one standard deviation below and above the mean</em> contains, approximately, 68% of the observations; the interval between <em>two standard deviations below and above the mean</em> contains, approximately, 95% of the observations; and the interval between <em>three standard deviations</em> below and above the mean contains, approximately, 99.7% of the observations, we could have concluded that 2.5 % of the manufacturer's grade A batteries will last <em>less</em> than 49 months, and, as a result, 1 - 0.025 = 0.975 or 97.5% will last more than 49 months.

We can conclude that with a less precise answer (but faster) because of the <em>symmetry of the normal distribution</em>, that is, 1 - 0.95 = 0.05. At both extremes we have 0.05/2 = 0.025 or 2.5% and we were asked for P(x>49) = P(z>-2) (see the graph below).

6 0

4 years ago

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

3 years ago