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ivann1987 [24]
3 years ago
15

87 markes out of 100 as a percentage

Mathematics
1 answer:
Maru [420]3 years ago
5 0
= (87/100*100)% =(8700/100)% =87%
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Simplify the expression to a polynomial in standard form. (X+1)(3x^2+9x+2)
OLEGan [10]

Answer:

3x^3 + 12x^2 +11x +2

Step-by-step explanation:

this is also a cube root expanded If that helps

4 0
2 years ago
5TH GRADE MATH! HELP! What is written in standard form? A. 0.8 B. 0.871 C. 871 D. 8,710
vladimir1956 [14]

Answer:B

Step-by-step explanation:

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3 years ago
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8.804 divided by 0.903 (include remainder) plz
nignag [31]

the answer is 9.74972315 I hope you have the right answer I hope this is helpful


7 0
3 years ago
Answer asap
balandron [24]

Answer:

A

Step-by-step explanation:

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3 years ago
For a given IQ test, an individual is considered a genius if their score falls more than three standard deviations from the mean
ki77a [65]

Answer:

P(Z>3) = 1-P(Z

So then the probability that an individual present and IQ higher than 3 deviation from the mean is 0.00135

And if we find the number of individuals that can be considered as genius we got: 0.00135*1500=2.025

And we can say that the answer is a.2

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu=?,\sigma=?)  

We are interested on this probability

P(X>X+3\mu)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

And we can find the following probablity:

P(Z>3) = 1-P(Z

So then the probability that an individual present and IQ higher than 3 deviation from the mean is 0.00135

And if we find the number of individuals that can be considered as genius we got: 0.00135*1500=2.025

And we can say that the answer is a/2.0

3 0
3 years ago
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