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2 years ago

I need the answers to all of them pleaseeee

Mathematics

2 answers:

irina1246 [14]2 years ago

6 0

Answer:

I think it is 43167 but am not really sure

fgiga [73]2 years ago

6 0

Answer:

HOW DID YOU DO THAT? BECAUSE I CANT DO THAT 3 PIC

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A RECTANGULAR PRISM HAS A VOLUME OF 96 FT. THE AREA OF THE BASE IS 24 FT. WHAT IS THE HEIGHT OF THE PRISM?

Reptile [31]

96
i hope it is right answer ;)

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3 years ago

PLEASE HELP I've been struggling on this one :(

In-s [12.5K]

Answer: The answer is (x+2)(x-4)(x-4)
Or as a short form you can write it as $(x+2) (x-4)^{2}$

Step-by-step explanation:

Answer is very lengthy but I am 100% sure that's the correct answer

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1 year ago

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number between 1 and 10, inclusive, is randomly chosen. Events A and B are defined as follows. A: {The number is even} B: {The n

Rufina [12.5K]

Answer:

A U B = {1,2,3,4,5,6,8}

Step-by-step explanation:

Let's define the sets.

Let ¢ = universal set ( Contains all the elements in the set)

¢ = {1, 2, 3, 4, 5, 6, 7, 8, 9 , 10}

A= {2, 4, 6, 8}

B= {1,2,3,4,5,6}

Therefore: A U B (A union B):

A U B = {1, 2, 3, 4, 5, 6, 8}

6 0

3 years ago

What is the exact volume of the sphere? 36.868π m³ 94.46π m³ 147.456π m³ 463.01π m³ Sphere with dotted line extending from middl

Bezzdna [24]

80% sure it b 94.46π m³

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2 years ago

Read 2 more answers

The probability of an event - picking teams. About (a) 10 kids are randomly grouped into an A team with five kids and a B team w

LuckyWell [14K]

Answer: The required probability is $\dfrac{4}{9}$

Step-by-step explanation:

Since we have given that

Number of kids = 10

Number of kids in Team A = 5

Number of kids in Team B = 5

There are three kids in the group, Alex and his two best friends Jose and Carl.

So, number of favourable outcome is given by

$2(\dfrac{8!}{3!\times 5!})$

Total number of outcomes is given by

$\dfrac{10!}{5!\times 5!}$

So, the probability that Alex ends up on the same team with at least one of his two best friends is given by

$\dfrac{2(\dfrac{8!}{5!\times 3!)}}{\dfrac{10!}{5!\times 5!}}\\\\=2\times \dfrac{8!}{3!}\times \dfrac{5!}{10!}\\\\=\dfrac{4}{9}$

Hence, the required probability is $\dfrac{4}{9}$

3 0

2 years ago