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2 years ago

In circle p, st = 30 ft. what is the circumference of the circle? 45p ft 15p ft 10p ft 30p ft

1 answer:

5 0

The answer to this question is:

In circle p, st = 30 ft. what is the circumference of the circle?
"30p ft"

Hope this helped, Taylenlane7544
Your Welcome :)

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If an exponential model was used to fit the data set below, which of the following would be the best prediction for the output o

satela [25.4K]

Answer:

The equation is found to be: $y = 50.6e^{0.16x}$

y(20) = 1241.34

Step-by-step explanation:

The given data is:

x: 3 7 11 14 17

y: 83 142 301 450 722

Now, we find sum summation values, relevant to the formula of exponential regression model, using calculator:

∑ ln y = 27.77305, ∑x ln y = 308.1494, ∑x = 52, ∑ x² = 664

and, n = no. of data points = 5

Now, we use formulae of exponential regression model to find out values of constant:

b = (n∑x lny - ∑x ∑ln y)/[n∑x² - (∑x)²]

b = [(5)(308.1494) - (52)(27.77305)]/[(5)(664) - (52)²]

b = 0.16

Now, for a;

a = (∑ln y - b∑x)/n

Therefore,

a = [(27.77305) - (0.16)(52)]/5

a = 3.9

For, α:

α = e^a = e^3.9

α = 50.6

So, the final equation of exponential regression model is given as:

$y = \alpha e^{bx}\\ y = 50.6e^{0.16x}$

Now, we find value of y for x = 20:

y(20) = (50.6) e^(0.16*20)

y(20) = 1241.34

8 0

2 years ago

Use a table of numerical values of f(x,y) for (x,y) near the origin to make a conjecture about the value of the limit of f(x,y)

grigory [225]

Seems to be that the limit to compute is

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+2y^2}$

Consider an arbitrary line through the origin $y=mx$ , so that we rewrite the above as

$\displaystyle\lim_{x\to0}\frac{mx^2}{x^2+2m^2x^2}=\lim_{x\to0}\frac m{1+2m^2}=\frac m{1+2m^2}$

The value of the limit then depends on the slope $m$ of the line chosen, which means the limit is path-dependent and thus does not exist.

8 0

2 years ago

Read 2 more answers

The mapping diagram shows a functional relationship

Gre4nikov [31]

Answer:

$f(4)=\frac{1}{2}$

f(x) = 4 when x is 8

Step-by-step explanation:

Domain is the set of x values that make the function defined. Allowed x values for the function (mapping).

The Range is the set of y values that make the function defined. Allowed y values for the function (mapping).

Whenever we need to find f(a), suppose, then we look for "a" in the domain and see its corresponding value mapping in the range.
Whenever we will be given a value for f(x) = a, suppose, and we have to find "x", we look at the value a in the range and find corresponding x value in the domain.

Firstly, we need f(4), so we look for "4" in domain and see which number it corresponds to in range.

That is $\frac{1}{2}$

Thus,

$f(4)=\frac{1}{2}$

Next,

We want "x" value that gives us a "y" value of 4. We look for "4" in the range and see which value it corresponds to. That is "8". So,

f(8) = 4

8 0

2 years ago

A semicircular flower bed has a diameter of 3 metres. What is the area of the flower bed?

Yuri [45]

The area of circle is

$A_{\text{circle}}=\pi r^2.$

Then the area of semicircular region is

$A_{\text{semicircular region}}=\dfrac{1}{2} \pi r^2.$

If r=3 m, you have

$A_{\text{semicircular region}}=\dfrac{1}{2} \pi \cdot 3^2=\dfrac{9\pi}{2}=4.5\pi \approx 14.13$ sq. m.

Answer: $A_{\text{semicircular region}}=4.5\pi \approx 14.13$ sq. m.

3 0

2 years ago

Read 2 more answers

Below are boxplots that summarize the weights (in pounds) of large samples from two breeds of dog: the Anatolian Shepherd and th

koban [17]

Answer:

(a) The median of both plots is same, 120 pounds is the median weight of Anatolian Shepherd and the Black Russian Terrier breed.

The range of both plots is also same that is 75 pounds.

The box plot of Anatolian Shepherd is skewed towards left therefore, it is considered to be positively distributed.

The box plot of Black Russian Terrier is skewed towards right therefore, it is considered to be negatively

(b) Any data point which is greater than 155 or smaller than 80 will be considered an outlier.

Step-by-step explanation:

A box plot is a graph which shows five statistical characteristics of a data set.

1. Maximum value

2. Minimum value

3. Median

4. Upper Interquartile

5. Lower interquartile

(a) Compare the distributions of weights for the two dog breeds

Please refer to the attached diagram of the question.

The median of both plots is same, 120 pounds is the median weight of Anatolian Shepherd and the Black Russian Terrier breed.

The range of Anatolian Shepherd is,

Range = Maximum value - Minimum value

Range = 175 - 100 = 75 pounds

The range of Black Russian Terrier is,

Range = Maximum value - Minimum value

Range = 155 - 80 = 75 pounds

Therefore, the range of both plots is also same that is 75 pounds.

A box-plot is considered to be normally distributed when the median is at the center of upper quartile and lower quartile.

The box plot of Anatolian Shepherd is skewed towards left therefore, it is considered to be positively distributed.

The box plot of Black Russian Terrier is skewed towards right therefore, it is considered to be negatively distributed.

(b) What weights would a Black Russian Terrier have to be to be considered an outlier?

An outlier is a data point in the data set that is very different from the other data points.

In this case, the maximum and minimum values in the Black Russian Terrier box-plot are

Maximum = 155

Minimum = 80

Therefore, any data point which is greater than 155 or smaller than 80 will be considered an outlier.

5 0

2 years ago