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3 years ago

Diego is making a large mural. He draws a hexagon with a perimeter of 10.5 meters. Each side of the hexagon is the same length.

Mathematics

1 answer:

miss Akunina [59]3 years ago

7 0

Answer:

Part A: 6a = 10.5 and a = 1.75 meters.

Part B: $3.87.

Step-by-step explanation:

Part A: Diego draws a hexagon with a perimeter of 10.5 meters and each side of the hexagon is the same length.

Now, the hexagon has 6 sides and if each side is of length 'a' meters, then the equation that models the situation is

6a = 10.5 ......... (1) (Answer)

⇒ a = 1.75 meters. (Answer)

Part B:

The total cost of supplies to paint the mural is $38.70. Diego and 9 friends divide the cost equally.

Hence, the total cost will be divided into 10 persons.

Therefore, each person will get $\frac{38.70}{10} = 3.87$ dollars. (Answer)

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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.

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

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.

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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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