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3 years ago

13

paige and her family went tp the movies. they bought 4 tickets and paid $12 for popcorn. they spent $40. how much did each ticke

t cost?

2 answers:

IgorLugansk [536]3 years ago

7 0

THEIR FAMILY PAID $8

geniusboy [140]3 years ago

6 0

Hello!

First of all, we subtract the popcorn cost, 12, from 40, giving us 32. Now we divide by 4 tickets.

32/4=8

Therefore, each ticket costed $8.

I hope this helps!

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a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

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$P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633$

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The mean of this distribution is:

$\mu=np=10*0.6=6$

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

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$P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989$

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