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3 years ago

9

Didn't find this anywhere. it might help others too. please help

1 answer:

dusya [7]3 years ago

3 0

The answer is B! hope that helped

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3 years ago

Assume that when adults with smartphones are randomly selected, 46% use them in meetings or classes. If 9 adult smartphone use

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Answer:

0.18173219.

Step-by-step explanation:

We have been asked to find what will be the probability that at least 6 of 9 adults use their smartphones in meetings or classes.

We will find our answer using Bernoulli's trails.

$_{r}^{n}\textrm{c}\cdot p^{r}\cdot (1-p)^{n-r}$

First of all we will find the probabilities when r is 6, 7,8 and 9 then we will add them all.

When r=6,

$_{6}^{9}\textrm{c}\cdot 0.46^{6}\cdot (1-0.46)^{9-6}$

$\frac{9!}{6!3!} *0.46^{6} *0.54^{3}$

$\frac{9*8*7*6!}{6!*3*2*1} *0.009474296896*0.157464$

$84*0.009474296896*0.157464=0.12532$

Similarly we will find Probabilities when r=7, 8 and 9.

When r=7

$_{7}^{9}\textrm{c}\cdot 0.46^{7}\cdot (1-0.46)^{9-7}$

$\frac{9!}{7!2!} *0.46^{7} *0.54^{2}$

$\frac{9*8*7!}{7!*2*1} *0.00435817657216*0.2916$

$36*0.00435817657216*0.2916=0.04575039$

When r=8

$_{8}^{9}\textrm{c}\cdot 0.46^{8}\cdot (1-0.46)^{9-8}$

$\frac{9!}{8!1!} *0.46^{8} *0.54$

$\frac{9*8!}{8!*1!} *0.00200476*0.54$

$9 *0.00200476*0.54=0.0097431336$

When r=9,

$_{9}^{9}\textrm{c}\cdot 0.46^{9}\cdot (1-0.46)^{9-9}$

$\frac{9!}{9!0!} *0.46^{9} *1$

$1 *0.00092219*1=0.00092219$

Now let us add all the probabilities to get the final answer.

$0.12532+0.04575+0.00974+0.00092219=0.18173219$

Therefore, probability that at least 6 of 9 adults use their smartphones in meetings or classes is 0.18173219.

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