Freshman Algebra help please anyone (just the answer no explanation)

Josh is going to the store to buy cakes to share with the fifth grade. He wants to share the cakes with 86 students. He buys 35

Ipatiy [6.2K]

Answer:

Everyone will get $\frac{35}{86}$ amount of cakes.

Step-by-step explanation:

Given:

Number of Cakes Josh buys = 35

Number of Students Josh wants to share = 86

Josh need to share cakes with fifth grade students.

We need to find the fraction of cakes everyone will get.

So to find the the fraction of cakes everyone will get can be calculated by dividing Number of Cakes Josh buys with Number of students he want to share with.

Amount of cakes Every one will get = $\frac{\textrm{Number of Cakes}}{\textrm{Number of Students}}$

Now Substituting the values we get;

Amount of cakes Every one will get = $\frac{35}{86}$

Hence Everyone will get $\frac{35}{86}$ amount of cakes.

7 0

3 years ago

Sandy works at a clothing store. She makes $9 per hour plus earns 10% commission on her sales. She worked 80 hours over the last

irina [24]

D. $720 is the answer

8 0

3 years ago

Can someone help please

maw [93]

Answer:

A, B, C, D, and F are all solutions.

Step-by-step explanation:

All of the possible answers are correct except for E. Answer E is 0.001 less than -4 which would not make the equation true. Hope this helps!!

6 0

3 years ago

What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success p? C

charle [14.2K]

Answer:

$(D)E[ X ] =np.$

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success p,

$f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq x\leq n$

$E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}$

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

$E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}$

Now,

$x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)$