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Arturiano [62]
3 years ago
8

Solve the equation. type the integer with it 32=8y

Mathematics
1 answer:
Sergio [31]3 years ago
3 0
 32 = 8y ⇔ y = 32 / 8 
              ⇔ y = 4
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Answer:

b

Step-by-step explanation:

because logic sence

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2 years ago
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Determine which numbers are solutions for x&gt;7 from the solution set<br> of { 7,8,9}
Vitek1552 [10]

Answer:

8, 9

Step-by-step explanation:

Since x > 7

This means that x cannot equal 7 but must be greater values than 7.

Thus 8 and 9 are solutions

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3 years ago
Who knows the area of this ? #struggleatmath
photoshop1234 [79]
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8 0
3 years ago
The new ointment was applied to four locations, and a control was applied to the other four. How many different choices were the
lubasha [3.4K]

Answer:

there are 70 possible choices for the four locations to apply the new ointment

Step-by-step explanation:

Since we have a total of 8 locations ( 4 to the new ointment and 4 to the control) ,  each one can be chosen and since the order of the locations that are chosen for the new ointment is not relevant  , then we know that the number of choices is given by the number of combinations of 4 elements in 8

number of combinations = 8 possible locations to the first ointment * 7 possible locations to the second ( since the first one was already located) * 6 to the third * 5 locations for the fourth / number of times the same combination is repeated ( the same locations but in different positions) = 8*7*6*5 / (4 possible positions for the first ointment* 3 possible positions to the second ointment (since the first one was already located * 2 possible positions of the third * 1 possible position of the fourth)

therefore

number of combinations = 8*7*6*5/(4*3*2*1 ) = 8!/((8-4)!*4!) = 70 possible combinations

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6 0
3 years ago
Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the nu
olga_2 [115]

Answer:

a

The probability that the selected joint was judged to be defective by neither of the two inspectors is   P(A' n B' ) = 0.8855

b

The probability that the selected joint was judged to be defective by inspector B but not by inspector A  is  P(A' n B) =0.0403

Step-by-step explanation:

From the question we are told that

   The sample size is n_s =  10000

    The number of outcome for inspector A is  n__{A}} = 742

    The number of outcome for inspector B is  n__{B}} = 745

     The number of joints judged defective by both inspector is n(A u B) =  1145

The the probability that the selected joint was judged to be defective by neither of the two inspectors is mathematically represented as

      P(A' n B' ) =  1 - P(A u B)

Now

       P(A\ u \ B) = \frac{n(Au B)}{n_s }

substituting values

        P(A\ u \ B) = \frac{1145}{ 10 000 }

So  

      P(A' n B' ) =  1 - \frac{1145}{10 000}

     P(A' n B' ) = 0.8855

the probability that the selected joint was judged to be defective by inspector B but not by inspector A  is mathematically represented as

     P(A' n B) =  P(A \ u \ B) -P(A)

Now

        P(A) =  \frac{n__{A}}{n_s}

substituting values

       P(A) =  \frac{742}{10 000}

So

     P(A' n B) =   \frac{1145}{10 000}  - \frac{742}{10 000}

    P(A' n B) =0.0403

7 0
3 years ago
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