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Yakvenalex [24]
3 years ago
9

Explain why the slope of a vertical line does not exist.

Mathematics
2 answers:
sattari [20]3 years ago
6 0
Lines<span> that are </span>vertical<span> have no </span>slope<span> (it </span>does not exist<span>). </span>Vertical lines<span> have "rise", but no "run". The rise/run formula for </span>slope<span> always has a zero denominator and is undefined. The equations for these </span>lines describe<span> what is happening to the x-coordinates.</span>
RSB [31]3 years ago
4 0
Because lines that are vertical are straight.
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Answer:

Let X be the event of feeling secure after saving $50,000,

Given,

The probability of feeling secure after saving $50,000, p = 40 % = 0.4,

So, the probability of not  feeling secure after saving $50,000, q = 1 - p = 0.6,

Since, the binomial distribution formula,

P(x=r)=^nC_r p^r q^{n-r}

Where, ^nC_r=\frac{n!}{r!(n-r)!}

If 8 households choose randomly,

That is, n = 8

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P(X=5)=^8C_5 (0.4)^5 (0.6)^{8-5}

=56(0.4)^5 (0.6)^3

=0.12386304

(b) the probability of the number that say they would feel secure is more than five

P(X>5) = P(X=6)+ P(X=7) + P(X=8)

=^8C_6 (0.4)^6 (0.6)^{8-6}+^8C_7 (0.4)^7 (0.6)^{8-7}+^8C_8 (0.4)^8 (0.6)^{8-8}

=28(0.4)^6 (0.6)^2 +8(0.4)^7(0.6)+(0.4)^8

=0.04980736

(c) the probability of the number that say they would feel secure is at most five

P(X\leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)

=^8C_0 (0.4)^0(0.6)^{8-0}+^8C_1(0.4)^1(0.6)^{8-1}+^8C_2 (0.4)^2 (0.6)^{8-2}+8C_3 (0.4)^3 (0.6)^{8-3}+8C_4 (0.4)^4 (0.6)^{8-4}+8C_5(0.4)^5 (0.6)^{8-5}

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=0.95019264

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