All categories

2 years ago

Can i get some help with this?

Mathematics

1 answer:

liberstina [14]2 years ago

7 0

Answer:

Look at the image, pls give brainliest!

Step-by-step explanation:

You might be interested in

The radius of a circle is 8 feet. What is the area of sector bounded by a 90° arc?

Lera25 [3.4K]

Since 90 degree is 1/4 of a circle, the area will be 1/4 th area of the full circle.

The radius is given by, r = 8 ft, therefore, the area of the full circle is,

$\begin{gathered} A=\pi\times r^2 \\ A=3.14\times8^2=200.96ft^2 \end{gathered}$

Therefore, the area of the sector is,

$A^{\prime}=\frac{1}{4}\times200.96ft^2=50.24ft^2$ $\begin{gathered} A^{\prime}=\frac{\theta}{360}\times\pi\times r^2 \\ \text{ =}\frac{\text{90}}{360}\times\pi\times8^2 \\ \text{ =}\frac{1}{4}\pi\times64=16\pi \end{gathered}$

4 0

1 year ago

Find the interquartile range of the data.

likoan [24]

The answer is 14. Here is how I work it out.

First we are going have to identify quartile 1 and quartile 3.

So after putting the numbers in order from least to greatest mark the number with a half way point.

This is optional but it will help us spot the sections better.

31,33,35,41,43,|46,48,49,49,50

Next put parentheses around the remaining groups of numbers.

(31,33,35,41,43,)|(46,48,49,49,50)

For the next step we have to find the median of each group.

(31,33,35,41,43,)|(46,48,49,49,50)

The median of each group, are called the quartiles, the median of the lower half is quartile 1, and the median of the upper half is quartile 3.

Now subtract quartile 1 from quartile 3.

49 – 35 = 14

So the interquartile range of this data set is 14.

4 0

3 years ago

Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq

Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

36 x² = 25

x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

$\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9$

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

$125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9$