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kodGreya [7K]
2 years ago
15

Divided it by 7 and added 25. the result was 34 what was my number?

Mathematics
2 answers:
AfilCa [17]2 years ago
6 0

Answer:63

Step-by-step explanation:

63/7+25=34

Aleksandr-060686 [28]2 years ago
6 0

Answer:

x = 63

Step-by-step explanation:

we know that...

(x will be the stand-in for "your number")

x ÷ 7 + 25 = 34

(x ÷ 7) + 25 = 34

To find x, we have to isolate x

first, we can subtract 25 from both sides (remember, if you do an action on one side, you must do the same action on the other side)

{why are we subtracting? we need to do the opposite of a sign to get rid of it. By subtracting, we are <em>undoing</em> addition; by multiplying we are undoing division (and vice versa)  }

(x ÷ 7) + 25 = 34

          - 25    - 25     (subtract 25)

x ÷ 7 = 9

  × 7    ×7       (multiplying by 7 to get rid of the "÷7")

x = 63

now, we can test out our x-value:

(63 ÷ 7)  +25 = 34

( 9 ) + 25 = 34

34 = 34

because this equation is true, we can be certain that x = 63

hope this helps!

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Step-by-step explanation:

Just write what you see by the letters

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Write the expression (4x-2).(6x 7) in the standard form of a anquadratic expression,ax^2 bx c what are the values of the coeffic
kipiarov [429]
Hello:
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6 0
3 years ago
A batch of 200 calculators contains 3 defective units. What the probability that a sample of three calculators will have/be
Anika [276]

Answer:

1. 0.048

2. 0.952

3. 0.6465

Step-by-step explanation:

Requirement 1

the probability of no defective calculator is 0.048

Given,

Total calculator = 200

Number of defective = 3

Probability of defective,  

p = 3/200

= 0.015

And the probability of non-defective,

q = 1 - p

=1-0.015

=0.985

The Probability distribution of defective follows the normal distribution.

P [X=0] = 〖200〗_(c_0 ) 〖(.015)〗^0 〖(.985)〗^(200-0)

P [X=0] = 0.048

Requirement 2

the probability of no defective is 0.048.

here, the probability of at least one defective means minimum 1 calculator can be defective. So the probability of at least one will be the probability of less than or equal 1.

P [X =less than or equal 1] = 1- P [X=0]

= 1 - 0.048  

= 0.952

So the probability of at least one defective is 0.952.

Requirement 3

c) the probability of all defective calculators is 0.6465.

P [X=3] = P[X=0]+ P[X=1]+ P[X=2]+ P[X=3]

Here,

P [X=0] = 0.048

P [X=1] = 〖200〗_(c_1 ) 〖(.015)〗^1 〖(.985)〗^(200-1)

= 0.148228

P [X=2] = 〖200〗_(c_2 ) 〖(.015)〗^2 〖(.985)〗^(200-2)

= 0.2245997

P [X=3] = 〖200〗_(c_3 ) 〖(.015)〗^3 〖(.985)〗^(200-3)

= 0.2257398

So,  P [X=3] = 0.048+0.148228+0.2245997+0.2257398  

= 0.6465675

So, the probability of all defective calculators is 0.6465.

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