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Nadusha1986 [10]
2 years ago
4

Find the distance between cities C and D to the nearest tenth

Mathematics
1 answer:
raketka [301]2 years ago
4 0

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===================================================

The formula to find the distance between two points ( x1 , y1 ) and ( x2 , y2) is:

d=\sqrt[]{(x2-x1)^2+(y2-y1)^2}

we need to find the distance between C and D

first, find the coordinates of the points

The coordinates of the point C = ( 3 , 6 )

the coordinates of the point D = ( 6 , 9 )

then use the formula of the distance:

d_{CD}=\sqrt[]{(6-3)^2+(9-6)^2}=\sqrt[]{3^2+3^2}=\sqrt[]{9+9}=\sqrt[]{18}=4.2426

so, the distance between C and D to the nearest tenth = 4.2

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What are the two numbers on the upper left and the three on the bottom?
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The computation of the equation shows.that x is -4 and y is 0.6.

<h3>How to calculate the value?</h3>

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The value of x and y will be calculated thus:

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8 0
2 years ago
So, I figured out X<br><br>so I just need help with what (x+17)°=<br><br>and what (3x-31)°​
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Opposite angles are same.

\\ \sf\longmapsto x+17=3x-31

\\ \sf\longmapsto 17+31=3x-x

\\ \sf\longmapsto 2x=48

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4 0
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A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the fil
MArishka [77]

Answer:

T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)

P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

X \sim N(12,0.3)  

Where \mu=12 and \sigma=0.3

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}

If we solve for the total T we got:

T= n \bar X

For this case then the expected value and variance are given by:

E(T) = n E(\bar X) =n \mu

Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2

And the deviation is just:

Sd(T) = \sqrt{n} \sigma

So then the distribution for the total would be also normal and given by:

T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)

And we want this probability:

P(T\leq 72)

And we can use the z score formula given by:

z = \frac{x-\mu}{\sigma}

P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z

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