We have to determine the probability that at least 2 of the dinners selected are pasta dinner. At least 2 of the dinners selected are pasta, it means that there can be 2 pasta, 3 , 4 or 5.

So, Probability with at least 2 of dinners

= $\frac{(^{12}C_{2} \times ^{10}C_{3})+ (^{12}C_{3} \times ^{10}C_{2})+(^{12}C_{4} \times ^{10}C_{1})+(^{12}C_{5})}{^{22}C_{5}}$

We will use the combination formula, which states $^nC_r = \frac{n!}{r! (n-r)!}$

= $\frac{(66 \times 120)+(220 \times 45)+(495 \times 10)+(792)}{26334}$

= $\frac{7920+9900+4950+792}{26334}$

= $\frac{23562}{26334}$

= 0.894

=0.90

Therefore, the probability that at least 2 of the dinners selected are pasta is 0.90

3 0

3 years ago

Please help!!!!!!!!!!

algol13

Answer:

2/25 of the dogs.

Step-by-step explanation:

Two of the dogs are fed only dry food, and there are 25 dogs, so the answer is 2/25.

3 0

3 years ago

Read 2 more answers

A new school has x day students and y boarding students.

guapka [62]

Given:

The fees for a day student are $600 a term.

The fees for a boarding student are $1200 a term.

The school needs at least $720000 a term.

To show:

That the given information can be written as $x + 2y\geq 1200$ .

Solution:

Let x be the number of day students and y be the number of boarding students.

The fees for a day student are $\$600$ a term.

So, the fees for $x$ day students are $\$600x$ a term.

The fees for a boarding student are $\$1200$ a term.

The fees for $y$ boarding student are $\$1200y$ a term.

Total fees for $x$ day students and $y$ boarding student is:

$\text{Total fees}=600x+1200y$

The school needs at least $720000 a term. It means, total fees must be greater than or equal to $720000.

$600x+1200y\geq 720000$

$600(x+2y)\geq 720000$

Divide both sides by 600.

$\dfrac{600(x+2y)}{600}\geq \dfrac{720000}{600}$

$x+2y\geq 1200$

Hence proved.

3 0

3 years ago