Here is a sequence of numbers:

Mathematics

1 answer:

Anettt [7]2 years ago

8 0

Triangular number sequnce

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PLEASE HELP ME THANKS IF u do

garik1379 [7]

Answer:

The third one

Step-by-step explanation:

It can't be the first one since the peak is 4 since it is the highest

It can't be the second one because the data is all different

It could be the third one since their are gaps

And It can't be the fourth one because the range is 7

So it must be the third one

6 0

2 years ago

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105 and a standard deviation sigma equa

Sindrei [870]

Answer:

0.7698

Step-by-step explanation:

If you call your random variable $X$ , then what you are looking for is

$P(87 \leq X \leq 123)$

because you want the probability of $X$ being <em>between 87 and 123.</em>

We need a table with of the normal distribution. But we can only find the table with $\mu = 0$ and $\sigma = 1$ . Because of that, first we need to <em>normalize </em>our random variable:

$Z = \frac{X - \mu}{\sigma} = \frac{X - 105}{15}$

(you can always normalize your variable following the same formula!)

now we can do something similar to our limits, to get a better expression:

$\frac{87 - 105}{15} = \frac{-18}{15} = -1.2$

$\frac{123 - 105}{15} = \frac{18}{15} = 1.2$

And we transform our problem to a simpler one:

$P(87 \leq X \leq 123) = P(-1.2 \leq Z \leq 1.2) = P(Z \leq 1.2) - P(Z \leq -1.2)$

(see Figure 1)

From our table we can see that $P(Z \leq 1.2) = 0.8849$ (this is represented in figure 2).

Remember that the whole area below the curve is exactly 1. So we can conclude that $P(Z \geq 1.2) = 0.1151$ (because 0.8849 + 0.1151 = 1). We also know the normal distribution is symmetric, then