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alisha [4.7K]
3 years ago
14

Which of the following sets could be the sides of a right triangle?

Mathematics
1 answer:
o-na [289]3 years ago
4 0

Answer:

1st answer

Step-by-step explanation:

From pythagorus theorem.

Since hypotenuse

=  \sqrt{  {2}^{2}  +  {3}^{2}  }=   \sqrt{ {13}}

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. A Gardener has 6203 plants. He wants to plant this in such a way that the number of rows and the number of columns remain the
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Answer:

38

Step-by-step explanation:

Since he wants to have equal number of rows and columns..the number of plants must be a root number. 6203 is not a root number. 6241 is the closest root number

(√6241= 79)

So he needs a minimum of 6241 plants in total for it.

6241-6203= 38

Therefore he needs 38 more plants

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3 years ago
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12a
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8 0
3 years ago
At one point the average price of regular unleaded gasoline was ​$3.39 per gallon. Assume that the standard deviation price per
irinina [24]

This question was not written completely

Complete Question

At one point the average price of regular unleaded gasoline was ​$3.39 per gallon. Assume that the standard deviation price per gallon is ​$0.07 per gallon and use​ Chebyshev's inequality to answer the following.

​(a) What percentage of gasoline stations had prices within 3 standard deviations of the​ mean?

​(b) What percentage of gasoline stations had prices within 2.5 standard deviations of the​ mean? What are the gasoline prices that are within 2.5 standard deviations of the​ mean?

​(c) What is the minimum percentage of gasoline stations that had prices between ​$3.11 and ​$3.67​?

Answer:

a) 88.89% lies with 3 standard deviations of the mean

b) i) 84% lies within 2.5 standard deviations of the mean

ii) the gasoline prices that are within 2.5 standard deviations of the​ mean is $3.215 and $3.565

c) 93.75%

Step-by-step explanation:

Chebyshev's theorem is shown below.

1) Chebyshev's theorem states for any k > 1, at least 1-1/k² of the data lies within k standard deviations of the mean.

As stated, the value of k must be greater than 1.

2) At least 75% or 3/4 of the data for a set of numbers lies within 2 standard deviations of the mean. The number could be greater.μ - 2σ and μ + 2σ.

3) At least 88.89% or 8/9 of a data set lies within 3 standard deviations of the mean.μ - 3σ and μ + 3σ.

4) At least 93.75% of a data set lies within 4 standard deviations of the mean.μ - 4σ and μ + 4σ.

​

(a) What percentage of gasoline stations had prices within 3 standard deviations of the​ mean?

We solve using the first rule of the theorem

1) Chebyshev's theorem states for any k > 1, at least 1-1/k² of the data lies within k standard deviations of the mean.

As stated, the value of k must be greater than 1.

Hence, k = 3

1 - 1/k²

= 1 - 1/3²

= 1 - 1/9

= 9 - 1/ 9

= 8/9

Therefore, the percentage of gasoline stations had prices within 3 standard deviations of the​ mean is 88.89%

​(b) What percentage of gasoline stations had prices within 2.5 standard deviations of the​ mean?

We solve using the first rule of the theorem

1) Chebyshev's theorem states for any k > 1, at least 1-1/k² of the data lies within k standard deviations of the mean.

As stated, the value of k must be greater than 1.

Hence, k = 3

1 - 1/k²

= 1 - 1/2.5²

= 1 - 1/6.25

= 6.25 - 1/ 6.25

= 5.25/6.25

We convert to percentage

= 5.25/6.25 × 100%

= 0.84 × 100%

= 84 %

Therefore, the percentage of gasoline stations had prices within 2.5 standard deviations of the​ mean is 84%

What are the gasoline prices that are within 2.5 standard deviations of the​ mean?

We have from the question, the mean =$3.39

Standard deviation = 0.07

μ - 2.5σ

$3.39 - 2.5 × 0.07

= $3.215

μ + 2.5σ

$3.39 + 2.5 × 0.07

= $3.565

Therefore, the gasoline prices that are within 2.5 standard deviations of the​ mean is $3.215 and $3.565

​(c) What is the minimum percentage of gasoline stations that had prices between ​$3.11 and ​$3.67​?

the mean =$3.39

Standard deviation = 0.07

Applying the 2nd rule

2) At least 75% or 3/4 of the data for a set of numbers lies within 2 standard deviations of the mean. The number could be greater.μ - 2σ and μ + 2σ.

the mean =$3.39

Standard deviation = 0.07

μ - 2σ and μ + 2σ.

$3.39 - 2 × 0.07 = $3.25

$3.39 + 2× 0.07 = $3.53

Applying the third rule

3) At least 88.89% or 8/9 of a data set lies within 3 standard deviations of the mean.μ - 3σ and μ + 3σ.

$3.39 - 3 × 0.07 = $3.18

$3.39 + 3 × 0.07 = $3.6

Applying the 4th rule

4) At least 93.75% of a data set lies within 4 standard deviations of the mean.μ - 4σ and μ + 4σ.

$3.39 - 4 × 0.07 = $3.11

$3.39 + 4 × 0.07 = $3.67

Therefore, from the above calculation we can see that the minimum percentage of gasoline stations that had prices between ​$3.11 and ​$3.67​ corresponds to at least 93.75% of a data set because it lies within 4 standard deviations of the mean.

4 0
3 years ago
Consider an experiment where two 6-sided dice are rolled. We can describe the ordered sample space as below where the first coor
agasfer [191]

Answer:

  • E = { (4,1) , (3,2) , (2,3) , (1,4) }
  • P(E)=\frac{1}{9}
  • P(F|E)=\frac{1}{4}

Step-by-step explanation:

Let's start writing the sample space for this experiment :

S= { (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) }

Let's also define the event E ⇒

E : '' The sum of the two dice is 5 ''

We can describe the event by listing all the favorables cases from S ⇒

E = { (4,1) , (3,2) , (2,3) , (1,4) }

In order to calculate P(E) we are going to divide all the cases favorables to E over the total cases from S. We can do this because all 36 of these possible outcomes from S are equally likely. ⇒

P(E)=\frac{4}{36}=\frac{1}{9} ⇒

P(E)=\frac{1}{9}

Finally we are going to define the event F ⇒

F : '' The number of the first die is exactly 1 more than the number on the second die ''

⇒

F = { (2,1) , (3,2) , (4,3) , (5,4) , (6,5) }

Now given two events A and B ⇒

P ( A ∩ B ) = P(A,B)

We define the conditional probability as

P(A|B)=\frac{P(A,B)}{P(B)} with P(B)>0

We need to find P(F|E) therefore we can apply the conditional probability equation :

P(F|E)=\frac{P(F,E)}{P(E)}   (I)

We calculate P(E)=\frac{1}{9} at the beginning of the question. We only need P(F,E).

Looking at the sets E and F we find that (3,2) is the unique result which is in both sets. Therefore is 1 result over the 36 possible results. ⇒

P(F,E)=\frac{1}{36}

Replacing both probabilities calculated in (I) :

P(F|E)=\frac{P(F,E)}{P(E)}=\frac{\frac{1}{36}}{\frac{1}{9}}=\frac{1}{4}=0.25

We find out that P(F|E)=\frac{1}{4}=0.25

6 0
4 years ago
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