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Andreyy89
3 years ago
14

The mean of twenty numbers is 42. If four of the

Mathematics
1 answer:
ehidna [41]3 years ago
4 0

Answer:

mean of other sixteen numbers is 40

Step-by-step explanation:

let the total of 20 numbers be x

x/20 = 42

20×x/20 =42×20

x=840

let the total of four numbers be y

y/4=50

4×y/4 = 50×4

y=200

so as to get the difference of x - y

=840-200

=640

the average of sixteen numbers

=640/16

=40

therefore the mean of sixteen numbers is 40

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natima [27]

\dfrac{x^3+10x^2+13x+39}{x^2+2x+1}

x^3=x\cdot x^2, and x(x^2+2x+1)=x^3+2x^2+x. Subtracting this from the numerator gives a remainder of

(x^3+10x^2+13x+39)-(x^3+2x^2+x)=8x^2+12x+39

8x^2=8\cdot x^2, and 8(x^2+2x+1)=8x^2+16x+8. Subtracting this from the previous remainder gives a new remainder of

(8x^2+12x+39)-(8x^2+16x+8)=-4x+31

-84x is not a multiple of x^2, so we're done. Then

\dfrac{x^3+10x^2+13x+39}{x^2+2x+1}=x+8+\dfrac{-4x+31}{x^2+2x+1}

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3 years ago
What is the answer to 9.2 x 0.3​
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2.76

Step-by-step explanation:

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3 years ago
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HELP PLEASE ASAP!!! 50 points
neonofarm [45]

Answer:

(a+b,c)

Step-by-step explanation:

Note that the midpoint formula is:

(\frac{x_{1} +x_{2}}{2}, \frac{y_{1} +y_{2}}{2})

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It follows that:

(\frac{(2a+2b +0}{2}, \frac{2c+0}{2})\\(\frac{(2(a+b)}{2}, \frac{2c}{2})\\\\(a+b,c)


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2x+5=10<br> I need to find out x pls help​
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Answer:

5

Step-by-step explanation:

2x + 5 = 10

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7 0
3 years ago
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2. A statistics student plans to use a TI-84 Plus calculator on her final exam. From past experience, she estimates that there i
Anarel [89]

Answer:

  1. P(≥1 working) = 0.9936
  2. She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.

Step-by-step explanation:

1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...

... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936

2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.

If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.

This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.

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