The plot below shows the volume of vinegar used by each of 171717 students in their volcano experiments.

Basile [38]

Choice B: I’m good a math and

STEP BY Step EXPLAINATION: 45 is double 25 and if you double 3.49=6.98 and choise B shows 4.79 soooo it’s choise B boooom.

Pls mark me Brainlyist

5 0

3 years ago

andy picked 5 tulips.brianna picked 3 more tulips than andy , carley picked 16 tulips. how many times as many tulips did carley

tatyana61 [14]

Andy picked 5
brianna picked 3 more then andy...she picked 8
carley picked 16

carley picked 2 times as many tulips as brianne....because carley picked 16 and brianna picked 8

8 0

3 years ago

Read 2 more answers

Let X be the number of Heads in 10 fair coin tosses. (a) Find the conditional PMF of X, given that the first two tosses both lan

inn [45]

Answer:

Follows are the solution to the given point:

Step-by-step explanation:

For option A:

In the first point let z be the number of heads which is available on the first two trails of tosses so, the equation is:

$P(X=k | z=2 ) = \begin{pmatrix} 10-2\\ k-2\end{pmatrix} (\frac{1}{2})^{k-2} (\frac{1}{2})^{10-k}$

$= \begin{pmatrix} 8\\k-2\end{pmatrix} (\frac{1}{2})^{k-2} (\frac{1}{2})^{10-k} \ \ \ \ \ \ \ \ \ \ \\\\$

$k= 2,3..........10$ For option B:

$P(X=k | X \geq 2 ) = \sum^{10}_{i=2} \begin{pmatrix} 10-i\\ k-i\end{pmatrix} (\frac{1}{2})^{k-i} (\frac{1}{2})^{10-k}$

$k= 2,3, 4.........10$

8 0

3 years ago

Evaluate 5! + 2!. 60 122 5,040

ludmilkaskok [199]

! Means starting from that number multiply by every number down to 1.

5!= 5•4•3•2•1 = 20•6=120

2!= 2•1=2

120+2=122 :)

8 0

3 years ago

Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5

alex41 [277]

The smallest such number is 1055.

We want to find $x$ such that

$\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}$

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

$x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}$

$x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}$

$x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}$

Taking everything together, we end up with the system

$\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}$

Now the moduli are coprime and we can apply the CRT.

We start with

$x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5$

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2$

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

$x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3$

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

$x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5$

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

$x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7$

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since $2\cdot4\equiv8\equiv1\pmod7$ , the inverse of 2 is 4.

So, we have to adjust $x$ to

$x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845$

and from the CRT we find

$x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}$

so that the general solution $x=210n+5$ for all integers $n$ .

We want a 4 digit solution, so we want

$210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5$

which gives $x=210\cdot5+5=1055$ .

5 0

3 years ago