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2 years ago

5

I have green ribbon that is 9 inches long and red ribbon that is one tenth the length of the green ribbon. What is the length of

the red ribbon

1 answer:

Juli2301 [7.4K]2 years ago

3 0

9 inches long

9 divided by 10 is .9

multiply it by 10 (to get the length of the purple ribbon) and the answer is 9

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If 6/9 is equal to 10/R, what is R?

sladkih [1.3K]

Answer:

r = 15

Step-by-step explanation:

I don't really have an explanation, I'm just good at quick math,

I'm sorry if you wanted an explanation, I would explain but I am really bad at explaining math.

5 0

2 years ago

Find the distance from point A to the given line:<br> A(-1, 7), y=3x<br><br> How do I solve this?

kipiarov [429]

he formula of a distance between a point (a, b)

and a line Ax + By + C = 0:

$d=\dfrac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}$

We have the point A(-1, 7) → a = -1, b = 7

and the line y = 3x → 3x - y = 0 → A = 3, B = -1, C = 0.

Substitute:

$d=\dfrac{|(3)(-1)+(-1)(7)+0|}{\sqrt{3^2+(-1)^2}}=\dfrac{|-3-7|}{\sqrt{9+1}}=\dfrac{|-10|}{\sqrt{10}}=\dfrac{10\sqrt{10}}{\sqrt{10}\cdot\sqrt{10}}\\\\=\dfrac{10\sqrt{10}}{10}=\sqrt{10}$

<h3>Answer: d = √10</h3>

3 0

2 years ago

I need urgent help plzz the answer will be marked as brainliest

ludmilkaskok [199]

51cm^2
27cm^2
78cm^2

to find the area of compound shapes, you break them apart and find each shapes area then add it together.

to find the area of a square, use A=l * w

to find the area of a triangle, use, A= 1/2 b * h

6 0

2 years ago

The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for

Blababa [14]

$f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}$

a. 9:00 AM is the 60 minute mark:

$f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248$

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

$\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173$

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

$\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982$

The probability of doing so for at least 2 of 5 days is

$\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1$

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then the desired probability is

$F_X(30)-F_X(15)\approx0.950-0.777=0.173$

7 0

3 years ago

***Please Help As Soon As Possible***

Nastasia [14]

Hey, ok so 1) is A
2) is C
3) is B
4) is B
5) is A
6) is C
7) is B
8) is B
9) is A
10) is C
How i go this is you just take each number and plug it in like 10 all you had to do was take one of the numbers in the answers and put it in as m like this 426/3=142 and see if that works. I would use a calculator <span />

5 0

3 years ago