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2 years ago

Hey i’ll give brainliest please help

Mathematics

1 answer:

just olya [345]2 years ago

7 0

I believe the answer is the fourth option.

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4. From a faculty of six professors, six associate professors, ten assistant professors, and twelve instructors, a committee of

Degger [83]

Answer:

P=0.228

Step-by-step explanation:

We know that from a faculty of six professors, six associate professors, ten assistant professors, and twelve instructors, a committee of size six is to be selected.

Therefore, we have 34 people.

We calculate the number of possible combinations:

$C^{34}_6=\frac{34!}{6!(34-6)!}=1344904\\$

Of the 6 professors we choose 2, and of the other 28 people we choose 4.

We calculate the number of favorable combinations:

$C_2^6\cdot C_4^{28}=\frac{6!}{2!(6-2)!}\cdot \frac{28!}{4!(28-4)!}=15\cdot 20475=307125\\$

Therefore, the probability is:

$P=\frac{307125}{1344904}=0.228$

8 0

3 years ago

To the people who need points, here you go. Just put a random answer. I know that someone is going to be ungrateful and just ste

denis-greek [22]

Answer:

Thanks your amazing i hope you have a great day.

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

In the diagram, the radius of the outer circle is

JulijaS [17]

the value of x is 8 cm.

<u>Step-by-step explanation:</u>

Correct Question : In the diagram, the radius of the outer circle is 2x cm and the radius of the inside circle is 6 cm. The area of the shaded region is 220π cm2. What is the value of x? Enter your answer in the box.

We have ,

the area of a circle = πr²

the outer circle area = $\pi(2x)^2 =4\pi x^2$

the inside circle area = $\pi (6)^2= 36\pi$

According to Question,

the outer circle area - the inside circle area = he shaded region

⇒ $\pi (4x^2)-36\pi =220\pi$

⇒ $x^2-9 =55$

⇒ $x^2=64$

⇒ $x =\sqrt{64}=8$

Therefore , the value of x is 8 cm.

5 0

2 years ago

The price is three for $1.48 how much will one pepper cost?

mr Goodwill [35]

Answer:

$0.4933

Step-by-step explanation:

i guess

Just divided by 3 ;)

8 0

2 years ago

In the figure, ∠1 ≅ ∠2, ∠3 ≅ ∠4, and . You can justify the conclusion that ∠TKL ≅ ∠TLK by stating _____

VladimirAG [237]

From the question, we know that we will be looking at <3, <4, and angles TKL and TLK. That being said, since <3 is congruent to <4, that means that angles TKL and TLK, which are each supplementary to either angle 3 or 4, are congruent because, since angles 3 and 4 are congruent, they are congruent because the supplements of congruent angles are congruent.

3 0

2 years ago

Read 2 more answers