The side lengths 30, 41, and 70 can form a triangle.

1 answer:

3 0

A = 30
B = 41
C = 70

A < B + C
30 < 41+70 ==> 30 < 111 (TRUE)

B < A + C ==>
41 < 30 + 70 ==> 41 < 100 (TRUE)

C < A+B
70 < 30 + 41 ==> 70 < 71 (TRUE)

Yes, can!

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Assume that when adults with smartphones are randomly selected, 51% use them in meetings or classes. If 7 adult smartphone use

Morgarella [4.7K]

Answer:

$P(X \geq 3) = 1-P(X$

If we find the individual probabilities we gotL

$P(X=0)=(7C0)(0.51)^0 (1-0.51)^{7-0}=0.0068$

$P(X=1)=(7C1)(0.51)^1 (1-0.51)^{7-1}=0.0494$

$P(X=2)=(7C2)(0.51)^2 (1-0.51)^{7-2}=0.1543$

And replacing we got:

$P(X \geq 3) = 1- [0.0068 +0.0494 +0.1543]= 0.7895$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=7, p=0.51)$