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2 years ago

. Find the range of the data. 7,4, 12, 1, 8, 8, 4, 14, 13, 9, 11, 10, 2,7,5, 3, 1, 8,5,3

Mathematics

1 answer:

Burka [1]2 years ago

5 0

Answer:

Range: 13

Step-by-step explanation:

Population size:20

Lowest value: 1

Highest value: 14

Answer: 13

<em><u>Hope this helps.</u></em>

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For the following right triangle, find the side length x. Round your answer to the nearest hundredth.

wlad13 [49]

Answer:

4.80

Step-by-step explanation:

To find the solution to this problem we will use the Pythagoras theorem.

${c}^{2} = {a}^{2} - {b}^{2}$

$= \sqrt{ {12}^{2} - {11}^{2}} \\ = 4.7958... \\ = 4.8$

4 0

2 years ago

Mark and Don are planning to sell each other their marble collections at a garage sale. If Don has 1 more than 3 times the numbe

Volgvan

Answer: Mark has 28 marbles while Don has 85 marbles.

Step-by-step explanation:

Let x represent the number of marbles that Mark has.

Let y represent the number of marbles that Don has.

If Don has 1 more than 3 times the number of marbles Mark has, it means that

y = 3x + 1

The total number of marbles is 113. It means that

x + y = 113- - - - - - - - - - - - 1

Substituting y = 3x + 1 into equation 1, it becomes

x + 3x + 1 = 113

4x + 1 = 113

4x = 113 - 1 = 112

x = 112/4

x = 28

y = 3x + 1 = 3 × 28 + 1

y = 85

4 0

3 years ago

I need help with this question

AlexFokin [52]

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3 0

1 year ago

Use prime factorization to find the least common multiple of 54 and 36

irinina [24]

Answer: 108

Explanation:

GCF(36,54) = 18
LCM(36,54) = ( 36 × 54) / 18
LCM(36,54) = 1944 / 18
LCM(36,54) = 108

6 0

2 years ago

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$

$P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021$

8 0

2 years ago