Ask question

Ask question

All categories

4 years ago

7

Suppose you are given a square with an inscribed circle as shown below. if the area of the square is 144 m2 what is the area of

the circle?

1 answer:

dimaraw [331]4 years ago

4 0

The area of the circle should be 113.097m^2

You might be interested in

Could someone explain this to me?

Ilya [14]

This is a simple differentiation problem. Let's start by taking the derivative of both sides (with respect to x):
$5y^4\frac{dy}{dx}+6y^2x+6x^2y\frac{dy}{dx}+20x^3 = 0$

Simplify:
$\frac{dy}{dx} (5y^4+6x^2y) + 6y^2x+20x^3 = 0$

Solve for dy/dx:
$\frac{dy}{dx} = \frac{-6y^2x-20x^3}{5y^4+6x^2y}$

Now, plug in the given points:
$\frac{dy}{dx} = \frac{-6(2)^2(-1)-20(-1)^3}{5(2)^4+6(-1)^2(2)} = \frac{24+20}{80+12}$

Further simplification gives:
$\frac{dy}{dx}|_{(2,-1)} =\frac{44}{92} = \frac{11}{23}$

So, your answer is 11/23 or B.

7 0

4 years ago

1. A flower shop wishes to add the valuable Waimea orchid to its product list. They purchase a large shipment of bulbs from a su

AysviL [449]

Answer:

The concern of the florist that these bulbs are not Waimea orchids, but a similar appearing hybrid maybe true.

Step-by-step explanation:

A Chi-square goodness of fit test can be used to perform the hypothesis test.

The hypothesis is defined as:

<em>H</em>₀: There is no difference between the observed and expected value.

<em>Hₐ</em>: There is a difference between the observed and expected value.

The test statistic is:

$\chi^{2}=\sum \frac{(O-E)^{2}}{E}$

The total number of Waimea orchid bulbs is, 60.

7x + 5x + 4x + 4x = 60

20x = 60

x = 3

The expected number of Waimea orchid bulbs are:

Blue = 7 × 3 = 21

Red = 5 × 3 = 15

Violet = 4 × 3 = 12

Orange = 4 × 3 = 12

Consider the table provided.

The test statistic value is:

$\chi^{2}=\sum \frac{(O-E)^{2}}{E}=14.981$

The decision rule:

The null hypothesis will be rejected if $\chi^{2}\geq \chi^{2}_{\alpha, (k-1)}$ .

Compute the critical value as follows:

$\chi^{2}_{0.05, (4-1)}= \chi^{2}_{0.05, (3)}=7.815$

The test statistic value is 14.981.

$\chi^{2}=14.981> \chi^{2}_{\alpha, (k-1)}=7.815$

The null hypothesis will be rejected at 5% level of significance.

Conclusion:

As the null hypothesis was rejected it implies that there is a significant difference between the observed and expected value.

Thus, the concern of the florist that these bulbs are not Waimea orchids, but a similar appearing hybrid maybe true.

4 0

4 years ago

Here are the prices of five new pairs of hockey skates, in dollars: 59, 60, 65, 99, 175. The standard deviation is about $49. If

Phoenix [80]

Answer:

The standard deviation of the new data will be increased as compared to the previous standard deviation of the data.

Step-by-step explanation:

The prices are given to be : 59, 60, 65, 99, 175

Standard deviation = $49

Now, if we add or subtract any constant value to each of the terms then the standard deviation remains unchanged.

But, we add a new price in the given data that is $450

$\text{New standard deviation = }\sqrt{\frac{\sum (\bar{x}-x_i)^2}{n}}\\\\\bar{x}\text{ is the mean of data after adding 450 and}\\x_i\text{ are the price values of the sample data}$

Hence, Standard deviation is calculated to be 139.5

Therefore, the standard deviation of the new data will be increased as compared to the previous standard deviation of the data.

6 0

3 years ago

Read 2 more answers

Solve for x and y 100x + 50y = 200

Eva8 [605]

Answer:

x = 1 and y = 2

Step-by-step explanation:

because 100(1) + 50(2) = 100 + 100 = 200

8 0

3 years ago

Addition of large numbers<br> 9,13,09,540+<br> 7,48,91,730+<br> 2,48,65,865

Marizza181 [45]

<span>191,067,135 is the answer.</span>

4 0

4 years ago