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Zielflug [23.3K]
2 years ago
15

The mall is located 2.4 kilometers north of jays house. The caffe shop is located 0.7 kilometers east of the mall. What is the d

istance between jays hair and the coffe shop
Mathematics
1 answer:
Iteru [2.4K]2 years ago
4 0

The distance<u> between </u>Jays hair and the caffe shop is 2.5 km.

Since the mall is located 2.4 kilometers<u> north</u> of jays house and the caffe shop is located 0.7 kilometers <u>east </u>of the mall, both directions are perpendicular. So, the distance between Jays hair and the caffe shop forms the hypotenuse side of this right angled triangle.

So, using Pythagoras' theorem, we find the hypotenuse side of this triangle, L.

Pythagoras' theorem states that the sum of squares of the sides of a right angled triangle equal the square of the hypotenuse side.

So, L = √[(2.4 km)² + (0.7 km)²]

L = √[(5.76 km² + 0.49 km²]

L = √(6.25 km²)

L = 2.5 km

So, the distance<u> between </u>Jays hair and the caffe shop is 2.5 km.

Learn more about Pythagoras' theorem here:

brainly.com/question/8171420

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Which of the following shows 12⁄28 written in prime factored form to help in reducing the fraction to simplest form?
mixas84 [53]
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Explanation:
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3 0
3 years ago
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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
svetlana [45]

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.

P(x\leq7)=1-\sum_{k=8}^{10}P(k)=1-(P(8)+P(9)+P(10))\\\\\\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\leq 7)=1-(0.1209+0.0403+0.006)=1-0.1672=0.8328

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3 years ago
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Answer:

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-4a²b²+18a³-2b³

Hope this helps

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2 years ago
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