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rjkz [21]
3 years ago
11

What number makes 3 and 71 divisible by 9?

Mathematics
1 answer:
Snezhnost [94]3 years ago
4 0
3 only 
<span>7 only </span>
<span>1 only </span>
<span>1, 4, and 7 </span>
<span>do you mean which digit is added to 371 so that it is divisible by 3 </span>
<span>test for 3, sum of digits divisible by 3 </span>
<span>3 + 7 + 1 = 11 </span>
<span>if you add 3, 11 + 3 = 14 not divisible by 3 </span>
<span>if you add 7, 11+ 7 = 18 divisible by 3 </span>
<span>if you add 1, 11+ 1 = 12 divisible by 3 </span>
<span>if you add 1, 4, 7 divisible by 3 </span>
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AnnZ [28]

Answer:

c=-2

Step-by-step explanation:

We are given the function:

f(x) = -2c + cx - x^2

And that:

\displaystyle f^{-1} (5) = -1

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Recall that by definition of inverse functions:

\displaystyle \text{If } f(a) = b, \text{ then } f^{-1}(b) = a

So, since f⁻¹(5) = -1, then f(-1) = 5.

Substitute:

f(-1) = 5 = -2c + c(-1) - (-1)^2

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Combine like terms:

6 = -3c

And divide. Hence:

c = -2

In conclusion, the value of <em>c</em> is -2.

8 0
3 years ago
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Bas_tet [7]

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Step-by-step explanation:

<u>Explanation</u>:-

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p(x)    0.091    0.334          0.408      0.166

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p_{i} = P(x=x_{i}) = p(x_{i}) for I= 1,2,3......

If the numbers p(x_{i}) , I =1 ,2,3,.....  satisfy the two conditions

i) p(x_{i}) \geq 0 for all values of i

ii) ∑P(X=x) =1

Given data condition(i) satisfied p(x_{i}) \geq 0 for all values of i

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sum off all probabilities is not equal to one

<u>Conclusion</u>:-

The given data is not probability distribution

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