What is 21X x 24y+ 24X -14y

Mathematics

1 answer:

Elenna [48]3 years ago

3 0

Answer:

21x x 10y + 24X

Step-by-step explanation:

21X x 24y +24X -14y

collect the like terms

24y-14y=10y

different terms cant multiply each other hence X and y can't multiply each other

So your answer becomes

21X x 10y +24X

I'm not thaaaaaat sure of my answer so ask more i'm just trying

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What is <br> 2/3 * 6/7 + 5/8 / 1 3/4

stealth61 [152]

28/5ths which equally 595o

5 0

4 years ago

PLEASE may I get help, I'm so lost. It would mean a lot to me, please.

MrMuchimi

9514 1404 393

Answer:

80
40
10 to 20, inclusive

Step-by-step explanation:

1. The various goals can be met without filling the theater. This fact means there is a range of possibilities for each of the answers. However, we take the wording, "because of the demand for tickets ..." to mean demand is high and the theater will be sold out.

Since two children's tickets will be sold for each adult ticket sold, the number of children's tickets is 2/3 of the total.

children's tickets = 2/3 × 120 = 80

2. The remaining 40 tickets will be adult tickets.

40 adult tickets will be sold.

3. The total revenue must be at least $1800. If we allow 'd' tickets to be sold at a discount, then we can find the limits on d using the inequality ...

24(40-d) +24(0.75)d +12(80) ≥ 1800 . . . . revenue from ticket sales

-6d ≥ -120 . . . . . . . . . . . . . . . . . . . collect terms, subtract 1920

d ≤ 20 . . . . . . . . . . divide by -6

At least 10 and at most 20 adult tickets can be sold with a discount.

("At least 10" comes from the problem requirements.)

4 0

3 years ago

Suppose that the probability that any particle emitted by a radioactive material will penetrate a certain shield is 0.01. If 10

AnnyKZ [126]

Answer:

a)P=0.42

b) $n\geq 297$

Step-by-step explanation:

We have a binomial distribution, since the result of each experiment admits only two categories (success and failure) and the value of both possibilities is constant in all experiments. The probability of getting k successes in n trials is given by:

$P=\begin{pmatrix}n\\ k\end{pmatrix} p^k(1-p)^{n-k}=\frac{n!}{k!(n!-k!)}p^k(1-p)^{n-k}$

a) we have k=2, n=10 and p=0.01:

$P=\frac{10!}{2!(10!-2!)}0.01^2(1-0.01)^{10-2}\\P=\frac{10!}{2!*8!}0.01^2(0.99)^{8}\\P=45*0.01^2(0.99)^8=0.42$

b) We have, $1-(1-p)^n=P$ , Here P is the probability that at least one particle will penetrate the shield, this probabity has to be equal or greater than 0.95. Therefore, this will be equal to subtract from the total probability, the probability that the particles do not penetrate raised to the total number of particles.