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Tju [1.3M]
3 years ago
11

There are between 24 and 40 students in a class,

Mathematics
2 answers:
USPshnik [31]3 years ago
8 0

Answer:

There are 33 students

Step-by-step explanation:

Because 7:4 should be in multiple of 11

erica [24]3 years ago
8 0
Because 4:7 should be multiply by 11
You might be interested in
Rewrite the function f(x)=-2(x+2)^2+6 in the form f(x)=ax^2+bx+c
Romashka [77]

Answer:

f(x) = -2x² - 8x - 2

General Formulas and Concepts:

  • Order of Operations: BPEMDAS
  • Expand by FOIL (First Outside Inside Last)
  • Standard Form: f(x) = ax² + bx + c
  • Vertex Form: f(x) = a(bx + c)² + d

Step-by-step explanation:

<u>Step 1: Define function</u>

Vertex Form: f(x) = -2(x + 2)² + 6

<u>Step 2: Find Standard Form</u>

  1. Expand by FOILing:                         f(x) = -2(x² + 4x + 4) + 6
  2. Distribute -2:                                     f(x) = -2x² - 8x - 8 + 6
  3. Combine like terms (constants):     f(x) = -2x² - 8x - 2
8 0
3 years ago
PLSSSSSSS HELPPPPPPP I WILL GIVE BRAINLIESTTTTTTTTTT!!!!!!!!!!!!!!!!!!!!!PLSSSSSSS HELPPPPPPP I WILL GIVE BRAINLIESTTTTTTTTTT!!!
Veronika [31]

Answer:

48

Step-by-step explanation:

4 0
2 years ago
bob is having a cookout. he bought 6 liters of sellers water. how many full 1 pint of sellers water can he pour?
Natali [406]

Answer:

192 servings

Step-by-step explanation:

8 0
3 years ago
A square pyramid is shown. What is the surface area?
sweet-ann [11.9K]
We need a picture to answer this
7 0
3 years ago
How would i solve a problem like this?
SVETLANKA909090 [29]

The number of combinations of size k that you can make with n items is given by the so-called binomial coefficient,

\dbinom nk = \dfrac{n!}{k!(n-k)!}

• n! is the number of ways of permuting n items.

• (n-k)! is the number of ways of permuting all but k of the n items.

Dividing n! by (n-k)! then gives the number of ways of permuting only k of the total n items.

• k! is the number of ways of permuting k items.

Dividing \frac{n!}{(n-k)!} by k! then removes all those permutations which contain the same items. We call these combinations.

For this problem we only care about counting combinations.

There are

\dbinom 63 = \dfrac{6!}{3!(6-3)!} = 20

ways of selecting any 3 girls from the total 6 girls in the entire group of people.

There are

\dbinom 72 = \dfrac{7!}{2!(7-2)!} = 21

was of selecting any 2 boys from the total 7 boys.

Then there are

\dbinom 63 \dbinom 72 = 20\cdot21 = \boxed{420}

ways of choosing a committee of 5 people consisting of 3 girls and 2 boys.

If the next question were, "What is the probability that a committee of 5 randomly selected people consists of 3 girls and 2 boys?", then you would additionally need to compute the number of ways one can make a committee of 5 people from the total 13, which is

\dbinom{13}5 = \dfrac{13!}{5!(13-5)!} = 1287

Then the probability of selecting such a committee at random is

\dfrac{\binom 63 \binom72}{\binom{13}5} = \dfrac{420}{1287} = \dfrac{140}{429} \approx 0.3263

8 0
1 year ago
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