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3 years ago

Help me? Giving out brain

Mathematics

2 answers:

grigory [225]3 years ago

6 0

24 people were wearing white shirts

Paladinen [302]3 years ago

6 0

24 is the answer because it’s 1/4th

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Given the set of numbers, 2, 3, 4, 8, 10 with mean = 6 The Second moment about the mean for this set is?

Rufina [12.5K]

Answer:

Given,

2,3,4,8,10

ex=2+3+4+8+10+x

=27+x

mean=6

no of data=5

now,

mean = ex/n

or,6=27+x/5

or,5×6=27+x

or,30=27+x

or,30-27=x

3=x

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3 years ago

Dec 21 /... the Christmas star is gonna be so pretty bruv

Svetradugi [14.3K]

Answer:

Even tho it's not Christmas yet Merry Christmas

Step-by-step explanation:

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3 years ago

PLEASE HELP PICTURE IS SHOWN

Alik [6]

Use the pythagorean theorem. You should come out with C. Hope this helps!

6 0

3 years ago

In a neutral atom that has an atomic mass of 22 and atomic number of 12, how many neutrons does this element have?

lozanna [386]

Given :

In a neutral atom that has an atomic mass of 22 and atomic number of 12.

To Find :-

The number of neutrons .

Solution :-

As we know that,

n(neutrons) + n(protons) = A
n(neutrons) + 12 = 2 2
n(neutrons) = 22-12
n(neutrons) = 10

Hence the required answer is 10.

8 0

3 years ago

Read 2 more answers

On a multiple choice test, if you randomly guessed on three questions, then what is the probability you got at least one of them

snow_lady [41]

Answer:

Number of Questions =3

Probability of giving a correct answer

$=\frac{1}{3}$

Probability of giving two correct answers

$=\frac{2}{3}$

Probability of giving all correct answers

$=\frac{3}{3}\\\\=1$

Probability that at least one of them is correct

$=_{1}^{3}\textrm{C}\\\\=\frac{3!}{(3-1)! \times1!}\\\\=3 \text{ways}\\\\=\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \\\\=\frac{1}{27}$

Probability that two of them is correct

$=_{2}^{3}\textrm{C}\\\\=\frac{3!}{(3-2)! \times2!}\\\\=3 \text{ways}\\\\=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \\\\=\frac{8}{27}$

Probability that all of them is correct

$=_{3}^{3}\textrm{C}\\\\=\frac{3!}{(3-3)! \times3!}\\\\=1 \text{way}\\\\=1$

So, Required probability

$=\frac{1}{27} \times \frac{8}{27} \times 1\\\\=\frac{8}{729}$

6 0

3 years ago