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3 years ago

12

The spinner below is spun 20 times. It lands on red 6 times, yellow 2 times, green 8 times, and blue 4 times.

1 answer:

PolarNik [594]3 years ago

5 0

Answer:

I think the answer is 42%

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Mary invests £120000 in a savings account.the account pays 1.5% compound interest per year work out the value of her investment

iris [78.8K]

Answer:

120,027£

Step-by-step explanation:

A=P(1+r/100)^n

120,000(1+1.5/100)^2

=120,027£

hope it helps

5 0

2 years ago

A linear function that represents the number of hot dogs purchased at a football game is compared to a different linear function

Vsevolod [243]

Answer:

Key features: the functions can not be parallels, this means that their respective slopes can not be equal or multiple to the other

Step-by-step explanation:

Take for instance 2 linear functions, and

If they intersect in some point in the space, say that is

Then (x1,y1)=(x3,y3), (x2.y2)= (x3,y3)

So, we can compare the two functions and get the following result;

It tells us that if M=m, we get an error, meaning that the functions are in fact parallels and there is no way that they meet in some point.

hope this helps!

5 0

3 years ago

What is the domain of this exponential function? y=2x

trasher [3.6K]

Answer:

x∈R

Step-by-step explanation:

$y=2^{x}$

2>0

The domain is all real numbers.

x∈R

8 0

3 years ago

Pls answer this i really need it

jeka57 [31]

Answer:

8.15

Step-by-step explanation:

4 0

2 years ago

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

oksano4ka [1.4K]

Answer:

Required center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

Step-by-step explanation:

Given semcircles are,

$y=\sqrt{1-x^2}, y=\sqrt{16-x^2}$ whose radious are 1 and 4 respectively.

To find center of mass, $(\bar{x},\bar{y})$ , let density at any point is $\rho$ and distance from the origin is r be such that,

$\rho=\frac{k}{r}$ where k is a constant.

Mass of the lamina=m= $\int\int_{D}\rho dA$ where A is the total region and D is curves.

then,

$m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k$

Now, x-coordinate of center of mass is $\bar{y}=\frac{M_x}{m}$ . in polar coordinate $y=r\sin\theta$

$\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta$

$=3k\int_{0}^{\pi}\sin\theta d\theta$

$=3k\big[-\cos\theta\big]_{0}^{\pi}$

$=3k\big[-\cos\pi+\cos 0\big]$

$=6k$

Then, $\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}$

y-coordinate of center of mass is $\bar{x}=\frac{M_y}{m}$ . in polar coordinate $x=r\cos\theta$

$\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta$

$=3k\int_{0}^{\pi}\cos\theta d\theta$

$=3k\big[\sin\theta\big]_{0}^{\pi}$

$=3k\big[\sin\pi-\sin 0\big]$

$=0$

Then, $\bar{x}=\frac{M_y}{m}=0$

Hence center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

3 0

3 years ago