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3 years ago

HEY CAN ANYONE HELP ME OUT IN DIS PLS

Mathematics

1 answer:

dimulka [17.4K]3 years ago

7 0

The answer is C.
You have to subtract 13 from both sides.

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I made my own tricky math question. But I don't know if its tricky for you. Amelia has 1 notebook on her desk, she put her noteb

fomenos

Answer:

1 thing, her pouch

Step-by-step explanation:

you sayed that she took her notebook away and the only thing onher desk is her pouch

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3 years ago

Find the slope and the line between each pair of points

dalvyx [7]

Girl what are you talking about

4 0

2 years ago

Help please!!!!!!!!!!!!!

IRISSAK [1]

Answer:

√20,

√19,

√17,

√23

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Eleven subtracted from two times a number is -31.

erica [24]

Answer:

x=-10

Step-by-step explanation:

$2x-11=-31\\$

Add 11 on both sides.

$2x-11=-31\\2x=-20\\2x/2x=-20/2\\x=-10$

7 0

2 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