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andrezito [222]
3 years ago
11

Factored 6x2 – 31x - 30 ​

Mathematics
1 answer:
ELEN [110]3 years ago
3 0

Answer:

942

Step-by-step explanation:

I used the clculater

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it would be 4.06 because 6/100 is 0.06 so 4.06 is your answer.

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Daily low temperatures in Columbus, OH in January 2014 were approximately normally distributed with a mean of 15.45 and a standa
Alona [7]

Answer: 12.10

Step-by-step explanation:

Given : Mean : \mu = 15.45

Standard deviation : \sigma = 13.70

The formula to calculate the z-score :-

z=\dfrac{x-\mu}{\sigma}

For x= 5 degrees

z=\dfrac{5-15.45}{13.70}=-0.7627737226\approx-0.76

For x= 10 degrees

z=\dfrac{10-15.45}{13.70}=-0.397810218\approx-0.40

The P-value : P(-0.76

=0.3445783-0.2236273=0.120951\approx0.1210

In percent , 0.1210\times100=12.10\%

Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%

5 0
3 years ago
Help Please. Can't seem to understand how to do this. I have tried and watch over 5 videos. Can't seem to get it.​
Snezhnost [94]

Answer:

The solutions are the following:

  • z=2(cos(π6)+isin(π6))=√3+12i

  • z=2(cos(2π3)+isin(2π3))=−1+i√3

  • z=2(cos(7π6)+isin(7π6))=−√3−12i

  • z=2(cos(5π3)+isin(5π3))=1−i√3

<em>hope this helps!! :) --Siveth</em>

7 0
2 years ago
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