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S_A_V [24]
3 years ago
5

PLZZZ NEED HELP!!!!!!

Mathematics
1 answer:
Lelu [443]3 years ago
4 0
The correct answer will be c
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in an examination 60% exam in failed in maths 55% failed in English and 25 failed in both subjects if none of the exam in past i
masha68 [24]

Answer:

60% x 25

55 % x 25

10 in english

25 is total and rest will be math

so 25-10 = 15

15 students failed math

4 0
2 years ago
What is the name of the place value of 9 in 199
Sliva [168]

The name of place value of 9 in 199 is ones.

The second 9 has a value of tens.

The number 1 has a place value of hundreds.

8 0
3 years ago
Read 2 more answers
Describe each step taken to solve the equation. Then, check the solution to see if it is valid. If it is not a valid solution, e
TEA [102]

Answer:

Step-by-step explanation:

given,

a) √x + 6 = 4                                                    

to solve the above equation subtract both side with 6

    √x + 6 - 6  = 4 - 6                  

                √x  = -2                  

squaring both side

                (√x)² = (-2)²              

                       x = 4            

a) ∛x + 6 = 4                                            

to solve the above equation subtract both side with 6

    ∛x + 6 - 6  = 4 - 6

                ∛x  = -2

cubing  both side

                (√x)³ = (-2)³

                      x = -8

4 0
3 years ago
How many 1/8-inch pieces of ribbon can be cut from a piece of ribbon that is 3/4 inch long? It can't be a fraction, sadly :')
SIZIF [17.4K]

Answer:

6 pieces.

Step-by-step explanation:

3/4 divided by 1/8

= 3/4 / 1/8        

Invert the 1/8 and multiply:

= 3/4 * 8/1

= 24/4

= 6.

7 0
3 years ago
Read 2 more answers
Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they
Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276

Number of ways of choosing 3 students From all males

n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220

The probability that all are male of choosing '3' students

P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }

P(E) =  \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0
3 years ago
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