Hong has a rope that is 9 meters long. He cuts away a piece that is 3.28 meters long. How long is the remaining piece of rope?

I have an hour left to finish this exam plz help me qwq"

sergejj [24]

Answer:

B. 10 E. 20 C. 60

Step-by-step explanation:

s = 2f

t = 2(f + s)

(15 + 15 + 30 + f + s + t)/6 = 25 This equation and the equations above show how the variables relate to each other

f = 10 The first number is 10

s = 2(10)=20 The second number is 20

t = 2(10 + 20) = 60 The third number is 60

If these answers are correct, please make me Brainliest!

6 0

2 years ago

In the diagram, GB = 2x + 3..

zaharov [31]

Answer-

$\boxed{\boxed{GB=15\ units}}$

Solution-

From the attachment,

AD = AE, so FA is a median.

BD = BF, so BE is a median.

CF = CE, so DC is a median.

And G is the centroid.

From the properties of centroid, we know that

The centroid divides each median in a ratio of 2:1

So,

$\Rightarrow FG:AG=2:1$

$\Rightarrow \dfrac{FG}{AG}=\dfrac{2}{1}$

$\Rightarrow FG=2\times AG$

$\Rightarrow 5x=2\times (x+9)$

$\Rightarrow 5x=2x+18$

$\Rightarrow 3x=18$

$\Rightarrow x=6$

So, GB will be $2(6)+3=15$ units

5 0

3 years ago

Read 2 more answers

What is the power of 15 ------------------

miv72 [106K]

Answer:

If something is "to the power of 15" then it means it is multiplied by itself 14 times (one less than 15 as the original number is already to the power of 1).

An example:

So 2 to the power of 15, which we write as 215 = 2 * 2 * 2 * 2 etc... (so we have 15 written 2's in total meaning we will make 14 multiplications) is equal to 32,768.

4 0

3 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago

Find the area of a circle having a circumference of 231 pi units. Round to the nearest tenth.

loris [4]

Do you by any chance have a diagram of the circle or the r value?

7 0

2 years ago