3^0·10^(-4)=<br> a)-120<br> b)1<br> c)1/10,000

The volume for the above figure is?

lara [203]

The area of the circle is πr², ie. 144π.
The volume of the cilinder is 12×144π = 1728π cm³.

I vote for answer A.

7 0

2 years ago

Translate the shape A three squares right and one square up

jonny [76]

To better give a visual, I drew it out instead.

The attachment is the solution.

8 0

1 year ago

Read 2 more answers

A picture measuring 2.5 inch high by 6 in wide is to be enlarged so that the width is now 18 in how tall will the picture be

lutik1710 [3]

The picture will now be 7.5 inches high by 18 inches wide.

7 0

2 years ago

An industry representative claims that 10 percent of all satellite dish owners subscribe to at least one premium movie channel.

Greeley [361]

Answer: 1) 0.6561 2) 0.0037

Step-by-step explanation:

We use Binomial distribution here , where the probability of getting x success in n trials is given by :-

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

, where p =Probability of getting success in each trial.

As per given , we have

The probability that any satellite dish owners subscribe to at least one premium movie channel. : p=0.10

Sample size : n= 4

Let x denotes the number of dish owners in the sample subscribes to at least one premium movie channel.

1) The probability that none of the dish owners in the sample subscribes to at least one premium movie channel = $P(X=0)=^4C_0(0.10)^0(1-0.10)^{4}$

$=(1)(0.90)^4=0.6561$

∴ The probability that none of the dish owners in the sample subscribes to at least one premium movie channel is 0.6561.

2) The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.

= $P(X>2)=1-P(X\leq2)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\= 1-[0.6561+^4C_1(0.10)^1(0.90)^{3}+^4C_2(0.10)^2(0.90)^{2}]\\\\=1-[0.6561+(4)(0.0729)+\dfrac{4!}{2!2!}(0.0081)]\\\\=1-[0.6561+0.2916+0.0486]\\\\=1-0.9963=0.0037$

∴ The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is 0.0037.

8 0

3 years ago

Find the rectangular coordinates of the point with the polar coordinates ordered pair 7 comma 2 pi divided by 3.

Morgarella [4.7K]

Answer:

$\left(-\dfrac{7}{2},\dfrac{7\sqrt{3}}{2}\right)$ .

Step-by-step explanation:

The given point is

$\left(7,\dfrac{2\pi}{3}\right)$

We need to find the rectangular coordinates of the given point.

If a polar coordinate is $(r,\theta)$ , then

$x=r\cos theta$

$y=r\sin theta$

In the given point $\left(7,\dfrac{2\pi}{3}\right)$ ,

$r=7,\theta=\dfrac{2\pi}{3}$

Now,

$x=7\cos \dfrac{2\pi}{3}$

$x=7\cos \left(\pi-\dfrac{\pi}{3}\right)$

$x=-7\cos \left(\dfrac{\pi}{3}\right)$

$x=-7\left(\dfrac{1}{2}\right)$

$x=-\dfrac{7}{2}$

and,

$y=7\sin \dfrac{2\pi}{3}$

$y=7\sin \left(\pi-\dfrac{\pi}{3}\right)$

$y=7\sin \left(\dfrac{\pi}{3}\right)$

$y=7\left(\dfrac{\sqrt{3}}{2}\right)$

$y=\dfrac{7\sqrt{3}}{2}$

Therefore, the required point is $\left(-\dfrac{7}{2},\dfrac{7\sqrt{3}}{2}\right)$ .

3 0

3 years ago

3^0·10^(-4)= a)-120 b)1 c)1/10,000