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klio [65]
3 years ago
10

Help me with this question please

Mathematics
1 answer:
IRISSAK [1]3 years ago
5 0

We have consecutive odd numbers for the sides of a triangle, perimeter 201.  We could write an equation, but clearly the middle side is 201/3 = 67, so the other two sides are 65 and 69.

This is almost an equilateral triangle, angles around 60 degrees, so not obtuse, meaning one angle over 90 degrees.

Scalene means the sides are all different lengths; we didn't even have to solve the problem to know this is true.

Smallest side 61, false.

Largest side 69, true.

This last one doesn't make much sense to me.  What's a unit here?  Dilation of 1/3 means all the linear measures become 1/3 as big, including the perimeter.  It would decrease much more than 3 inches, since it would go from 201 inches to 67 inches.  If I have to answer I'd go with false.


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To decrease the impact on the environment, factory chimneys must be high enough to allow pollutants to dissipate over a larger a
Komok [63]

Answer:

The probability hat the sample mean height for the 40 chimneys is greater than 102 meters is 0.1469.

Step-by-step explanation:

Let the random variable <em>X</em> be defined as the height of chimneys in factories.

The mean height is, <em>μ</em> = 100 meters.

The standard deviation of heights is, <em>σ</em> = 12 meters.

It is provided that a random sample of <em>n</em> = 40 chimney heights is obtained.

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

Since the sample selected is quite large, i.e. <em>n</em> = 40 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean heights of chimneys.

\bar X\sim N(\mu_{\bar x},\ \sigma^{2}_{\bar x})

Compute the probability hat the sample mean height for the 40 chimneys is greater than 102 meters as follows:

P(\bar X>102)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}})>\frac{102-100}{12/\sqrt{40}})

                    =P(Z>1.05)\\=1-P(Z

*Use a <em>z</em>-table fr the probability.

Thus, the probability hat the sample mean height for the 40 chimneys is greater than 102 meters is 0.1469.

8 0
3 years ago
Graph f(x) =x and g(x) =-1/3x+2
Serjik [45]
Hello! I'll write the instructions to graph these functions.

f(x)=x
Technically, this function is y=x, so the slope would be 1. To graph this one, start at the origin (0,0) and move up one unit, and to the right one unit since this is a positive slope.

g(x)= -1/3x+2
First, plot a point at y=2 when x=0. 2 is your y-intercept. Your slope is negative, so the line will be decreasing. From your first point, head down 1 unit and to the right 3 units. Continue plotting points from the previous points.

Also, if you have a graphing calculator, here are the steps to graphing the functions: ON, Y= (enter your functions), and press GRAPH or 2nd, TABLE to see individual points. Hope this helps! :)
3 0
3 years ago
First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x
e-lub [12.9K]

Answer:

(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

\int x ln(5+x)dx

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

\int (U-5) ln(U)dU

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

\int (pq')=pq-\int qp'

so we must define p, q, p' and q':

p=ln U

p'=\frac{1}{U}dU

q=\frac{U^{2}}{2}-5U

q'=U-5

and now we plug these into the formula:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU

Which simplifies to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU

Which solves to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C

so we can substitute U back, so we get:

\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C

and now we can simplify:

\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

notice how all the constants were combined into one big constant C.

7 0
3 years ago
The height of two similar pails are 12cm and 8cm.The larger pail can hold 2litres. What is the capacity of the smaller pail?
Vedmedyk [2.9K]

Answer:

1.33 liters

Step-by-step explanation:

Volume of any shape is given by cross sectional area multiplied by height. Therefore, V=Ah

Where A is area and h is height

Since their cross sections are the same, then volume depends on height.

If 12cm equals 2 liters

8cm equals x

X=(2*8)/12=1.3333333333333 liters

Approximayely, the smaller pail has capacity of 1.33 liters

7 0
3 years ago
What equation could you use to solve the problem below? The math clubs car can travel 9 m in a second. The race track for the so
WARRIOR [948]

Answer:

D

Step-by-step explanation:

Because if you divide it becomes 3

then to check if its right you just need to multiply 3 and 9 and it becomes 27

5 0
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