Ask question

Ask question

All categories

2 years ago

13

Two students are inviting friends to a mutual party,each student is inviting 5 of their 15 mutual friends, what is the probabili

ty that they will invite the same 5 friends?
A. 1 in 360360
B. 1 in 3003
C. 1 in 4368
D. 1 in 8008
( please help :( this is a question from Plato )

1 answer:

Nesterboy [21]2 years ago

7 0

Answer:

D?

Step-by-step explanation:

You might be interested in

Can someone help me (3.7x10-8)(2.9x10⁶

Salsk061 [2.6K]

Scientific notation:
8.41 x 10^7

expanded form:
84100000

3 0

2 years ago

a chef uses 4 3/4 cups of broth for 10 servings of soup.how much broth is is one serving . let x represent amount. 4 3/4/10 =x/1

enyata [817]

Let's firstly convert the mixed fraction to "improper" and then proceed,

$\bf \begin{array}{ccll}
\stackrel{broth}{cups}&servings\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
4\frac{3}{4}&10\\\\
x&1
\end{array}\implies \cfrac{\quad \stackrel{mixed}{4\frac{3}{4}}\quad }{x}=\cfrac{10}{1}\implies \cfrac{\quad \frac{4\cdot 4+3}{4}\quad }{x}=\cfrac{10}{1}$

$\bf \cfrac{\quad \stackrel{improper}{\frac{19}{4}}\quad }{x}=\cfrac{10}{1}\implies \cfrac{\quad \frac{19}{4}\quad }{\frac{x}{1}}=\cfrac{10}{1}\implies \cfrac{19}{4}\cdot \cfrac{1}{x}=\cfrac{10}{1}\implies \cfrac{19}{4x}=\cfrac{10}{1}
\\\\\\
19=40x\implies \cfrac{19}{40}=x$

6 0

2 years ago

What is the equation for this math problem? Steven spent $90 on clothes. He bought a pair of jeans that cost $34 and spent the r

dexar [7]

Answer:

The equation for the given maths problem is $ 34 + $ 8x = $ 90

Step-by-step explanation:

According to question :

The Total amount spent on cloths = $ 90

The cost of a pair of jeans = $ 34

The cost of each t-shirt pair = $ 8

The amount spent on t-shirt = $ 90 - $ 34 = $ 56

Let the number of t-shirt pair = x

So, $ 34 + $ 8x = $ 90

Or, $ 8x = $ 90 - $ 34

∴ x = $\frac{ 90 - 34}{8}$

Hence The equation for the given maths problem is $ 34 + $ 8x = $ 90 Answer

8 0

2 years ago

Somebody pleaseeee help

aniked [119]

The area of the cylinder is 490.09 ft squared

8 0

2 years ago

Verify a(b-c)=ab-ac for a=1.6;b=1/-2;& c=-5/-7

harina [27]

Given:

$a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}$

To verify:

$a(b-c)=ab-ac$ for the given values.

Solution:

We have,

$a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}$

We need to verify $a(b-c)=ab-ac$ .

Taking left hand side, we get

$a(b-c)=1.6\left(\dfrac{1}{-2}-\dfrac{-5}{-7}\right)$

$a(b-c)=1.6\left(-\dfrac{1}{2}-\dfrac{5}{7}\right)$

Taking LCM, we get

$a(b-c)=1.6\left(\dfrac{-7-10}{14}\right)$

$a(b-c)=\dfrac{16}{10}\left(\dfrac{-17}{14}\right)$

$a(b-c)=\dfrac{8}{5}\left(\dfrac{-17}{14}\right)$

$a(b-c)=-\dfrac{68}{35}\right)$

Taking right hand side, we get

$ab-ac=1.6\times \dfrac{1}{-2}-1.6\times \dfrac{-5}{-7}$

$ab-ac=-\dfrac{16}{20}-\dfrac{8}{7}$

$ab-ac=-\dfrac{4}{5}-\dfrac{8}{7}$

Taking LCM, we get

$ab-ac=\dfrac{-28-40}{35}$

$ab-ac=\dfrac{-68}{35}$

Now,

$LHS=RHS$

Hence proved.

7 0

2 years ago