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Sindrei [870]
3 years ago
13

a math class has 12 female students. Out of a total of 20 students in class what is the probability of randomly selecting a male

student from the class
Mathematics
2 answers:
Likurg_2 [28]3 years ago
8 0
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Calculate the number of male students in the class:

20 - 12 = 8

The probability would be 8/20 which can be simplified to 2/5

(if the simplified answer is required, write the second option, otherwise, write the first)

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

Sauron [17]3 years ago
3 0

There are 20 students in total.

12 of them are female, meaning 8 of them are male.

8/20 students are male.

Simplify this: 2/5

Because both 8 and 20 are divisible by 4.

2/5 is 40%

There is a 40% of randomly selecting a male student.

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What’s the length of BD ?
Rina8888 [55]

Answer:

BD = 4.99

Step-by-step explanation:

You can simply use the trigonometric identity tangent to solve for length BD.

Tan = opposite/adjacent

In this case we have,

Tan 31 = 3/BD

BD = 3/Tan 31

BD = 4.99

3 0
3 years ago
Read 2 more answers
If f(1) = 2 – 2 anid 9(37)<br> and g(x) = x2 – 9, what is the domain of g(x) = f(x)?
lana66690 [7]

Answer:

B

Step-by-step explanation:

Let divide g(x) by f(x)

\frac{ {x}^{2} - 9 }{2 - x {}^{ \frac{1}{2} } }

The domain of a rational function cannot equal zero so let set the bottom function to zero.

2 - x {}^{ \frac{1}{2} }  = 0

x  {}^{ \frac{1}{2} }  = 2

Square both sides

x = 4

Also we can simplify the bottom denomiator into a square root function

2 -  \sqrt{x}

The domain of a square root function is all real number greater than or equal to zero because a square root of a negative number isn't graphable.

So we must find a answer that

  • Disincludes 4 from the interval
  • Doesnt range in the negative number or infinity)
  • Range out in positve infinity
  • The answer to that is B

8 0
3 years ago
I need help please :)
IRISSAK [1]

Answer:

Step-by-step explanation:

Just subtract all the numbers from 38.75

3 0
3 years ago
Add -46+(-48)=<br> -42+46=
dalvyx [7]
-46 + (-48) = -94

-42 + 46 = 4
4 0
3 years ago
If k‐th term of a sequence is ak = ( -1) ^k(2-k)k/2k-1
max2010maxim [7]

Given:

kth term of a sequence is

a_k=\dfrac{(-1)^k(2-k)k}{2k-1}

To find:

The next term a_{k+1} when k is even.

Solution:

We have,

a_k=\dfrac{(-1)^k(2-k)k}{2k-1}

Put k=k+1, to get the next term.

a_{k+1}=\dfrac{(-1)^{k+1}(2-(k+1))(k+1)}{2(k+1)-1}

If k is even, then k+1 must be odd and odd power of -1 gives -1.

a_{k+1}=\dfrac{(-1)(2-k-1)(k+1)}{2k+2-1}

a_{k+1}=\dfrac{(-1)(1-k)(1+k)}{2k+1}

a_{k+1}=-\dfrac{1^2-k^2}{2k+1}    [\because (a-b)(a+b)=a^2-b^2]

a_{k+1}=-\dfrac{1-k^2}{2k+1}

Therefore, the next term is a_{k+1}=-\dfrac{1-k^2}{2k+1}.

7 0
3 years ago
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