When given 36 cubes, list 4 ways rectangular prisms can be created with a volume of 36 cubic units

Jelaskan ciri-ciri lingkungan bersih tolong di jawab yaah

rusak2 [61]

Answer:

huh hi what is the question?????

Step-by-step explanation:

5 0

3 years ago

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Let f(x)= x^4 + ax^2 +b. The graph of f has a relative minimum at (0,1) and an inflection point when x=1. The values of a and b

Delvig [45]

$f(x)=x^4+ax^2+b\\f'(x)=4x^3+2ax\\f''(x)=12x^2+2a\\f''(1)=12(1)^2+2a\\12(1)=-2a\\a=-6\\\\f(x)=x^4-6x^2+b\\(1)=(0)^4-6(0)^2+b\\b=1$

a=-6
b=1

8 0

3 years ago

Which equations are equivalent to 1/5 + 2/3 |2-xl = 4/15? Check all that apply.

Kazeer [188]

Answer:

(2 - x ) = 1 / 10

Step-by-step explanation:

1 / 5 + 2 / 3 (2 - x ) = 4 / 15

Subtract 1 / 5 from both sides

2 / 3 (2 - x ) = 4 / 15 - 1 / 5

2 / 3 (2 - x ) = 1 / 15

Multiply 3 / 2 on both sides

(2 - x ) = 1 / 15 * 3 / 2

(2 - x ) = 1 / 10

4 0

3 years ago

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A business organization needs to make up a 7 member fund-raising committee. The organization has 6 accounting majors and 7 finan

photoshop1234 [79]

Answer: 658 ways.

Step-by-step explanation:

To find the number of ways the number "r" items can be chosen from the available number "n", the combination formula for selection is used. This formula is denoted as:

nCr = n! / (n-r)! × r!

Where n! = n×(n-1)×(n-2) ... ×3×2×1.

If we have 6 accounting majors and 7 finance majors and we are to choose a 7-member committee from these with at least 4 accounting majors on the committee, then the possibilities we have include:

[4 accounting majors and 3 finance majors] Or [5 accounting majors and 2 finance majors] or [ 6 accounting majors and 1 finance major].

Mathematically, this becomes:

[6C4 × 7C3] + [6C5 × 7C2] + [6C6×7C1]

525 + 126 + 7 = 658 ways.

Note: it is 6C4 because we are choosing 4 accounting majors from possible 6. This applies to other selection possibilities.

8 0

3 years ago

A multiple-choice test has 30 questions and each one has five possible answers, of which only one is correct. If all answers wer

kumpel [21]

Answer: the probability is 0.13

Step-by-step explanation:

First, each question has 5 options and only one is correct.

Then, selecting at random, the probability of getting a correct answer is equal to:

p = 1/5 = 0.20 (and the probability of getting it incorrect is p = 4/5 = 0.80)

Now, if out of 30 questions, we got 4 correct and 26 incorrect, the probability for a given combination is;

p = (0.20^4)*(0.80^26)

But we also need to multiply this by the total number of combinations.

This is we have 30 questions in total, and we can select 4 of them that will be the correct ones.

Now, if we have N objects in total, the number of different combinations of K elements out of those N elements is

$C = \frac{N!}{(N-K)!*K!}$

In this case, N = 30 and K = 4.

$C = \frac{30!}{(30 -4)!*4!} = \frac{30*29*28*27}{4*3*2} = 27,405$

Then the probability of getting exactly 4 correct answers is:

P = (0.20^4)*(0.80^26)*27,405 = 0.13

5 0

3 years ago