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olga nikolaevna [1]
3 years ago
6

At the school talent show, 1/3 of the people were boys, 3/6 were girls, and the rest were adults. If there were 50 more girls th

an adults, how many people were there in total? *
Mathematics
1 answer:
strojnjashka [21]3 years ago
3 0

Answer:

150 people

Step-by-step explanation:

Lets call the total amount of people at the talent show x. Since 3/6=1/2, we know that half the people are girls, 1/3 are boys. If we added 1/2 and 1/3 we get 5/6, 1-5/6=1/6, so now we know that 1/6 of the people are adults. Because it says "If there were 50 more girls than adults" so from that we can see that 1/2x =1/6x+50. If we were to simplify the equation, we 1/3x=50, so x=150

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A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain
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Answer:

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).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

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

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55

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.

LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8

The probability of having between 4 and 8 customers choosing the oversize version is:

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

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

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