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olga2289 [7]
3 years ago
10

Which method of gathering data is described by the statement "these are run only when probabilities are known as fact

Mathematics
1 answer:
HACTEHA [7]3 years ago
3 0
It seems that you have missed the necessary options for us to answer this question. But anyway, here is the answer. The method <span>of gathering data that is described by the statement "these are run only when probabilities are known as fact" is called SIMULATION. Hope this answers your question.</span>
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What is the image of the point (4,7) after a reflection about the line y = 4.?
lord [1]

Answer:

(4, 1)

Step-by-step explanation:

See attachment courtesy of Desmos.

8 0
2 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
1. Write an equation in slope-intercept form for a line passing through (-3,3) with a slope of 1.
PIT_PIT [208]

1)


\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{3})~\hspace{10em} slope = m\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-3=1[x-(-3)] \\\\\\ y-3=1(x+3)\implies y-3=x+3\implies y=x+6


2)


\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-2}{1-(-4)}\implies \cfrac{10}{1+4}\implies \cfrac{10}{5}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=2[x-(-4)]\implies y-2=2(x+4) \\\\\\ y-2=2x+8\implies y=2x+10

8 0
3 years ago
Help! If you know this can you tell me how to do it?
aleksandr82 [10.1K]

Answer:

c

Step-by-step explanation:

Here's how this works:

Get everything together into one fraction by finding the LCD and doing the math.  The LCD is sin(x) cos(x).  Multiplying that in to each term looks like this:

[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?

In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

\frac{sin^2(x)}{sin(x)cos(x)}+\frac{cos^2(x)}{sin(x)cos(x)}=?

Put everything over the common denominator now:

\frac{sin^2(x)+cos^2(x)}{sin(x)cos(x)}=?

Since sin^2(x)+cos^2(x)=1, we will make that substitution:

\frac{1}{sin(x)cos(x)}

We could separate that fraction into 2:

\frac{1}{sin(x)}×\frac{1}{cos(x)}

\frac{1}{sin(x)}=csc(x)  and  \frac{1}{cos(x)}=sec(x)

Therefore, the simplification is

sec(x)csc(x)

5 0
3 years ago
Helppppppp pleaseeeee
AleksandrR [38]

Answer:

is it asking how many times u would get then number 5 when u roll or asking how you would expect for the first 5 rolls?

4 0
1 year ago
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