3 years ago

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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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

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The P-value : $P(-0.76$

$=0.3445783-0.2236273=0.120951\approx0.1210$

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

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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>

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2 years ago

Factored 6x2 – 31x - 30 ​

Factored 6x2 – 31x - 30