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3 years ago

I NEED HELP WITHIN AN HOUR!!! HELP!!

2 answers:

7 0

In order to find the coffee shop location, you find the midpoint of the church and the school.
Church: (100, 50) and School: (200, 100)
Midpoint or Coffee shop: (150, 75)
x=(100+200)/2=150
y=(50+100)/2=75

Hatshy [7]3 years ago

4 0

(150, 75)

Hope this helps you out Elizabeth0915! :D

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Plzz help e for brainliest if 2 people answer

Doss [256]

Answer:

Step-by-step explanation:

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3 years ago

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Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side so proof is complete

8 0

3 years ago

Read 2 more answers

Lucy scored 65 points in a game. Eva scored e points in the same game

kondor19780726 [428]

I guess you need to find what E is, though sadly this isnt enough information to me....

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3 years ago

Test scores of the student in a school are normally distributed mean 85 standard deviation 3 points. What's the probability that

Mrrafil [7]

Answer:

The probability that a random selected student score is greater than 76 is $\\ P(x>76) = 0.99865$ .

Step-by-step explanation:

The Normally distributed data are described by the normal distribution. This distribution is determined by two parameters, the population mean $\\ \mu$ and the population standard deviation $\\ \sigma$ .

To determine probabilities for the normal distribution, we can use the standard normal distribution, whose parameters' values are $\\ \mu = 0$ and $\\ \sigma = 1$ . However, we need to "transform" the raw score, in this case x = 76, to a z-score. To achieve this we use the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

And for the latter, we have all the required information to obtain z. With this, we obtain a value that represent the distance from the population mean in standard deviations units.

<h3>The probability that a randomly selected student score is greater than 76</h3>

To obtain this probability, we can proceed as follows:

First: obtain the z-score for the raw score x = 76.

We know that:

$\\ \mu = 85$

$\\ \sigma = 3$

$\\ x = 76$

From equation [1], we have:

$\\ z = \frac{76 - 85}{3}$

Then

$\\ z = \frac{-9}{3}$

$\\ z = -3$

Second: Interpretation of the previous result.

In this case, the value is three (3) standard deviations below the population mean. In other words, the standard value for x = 76 is z = -3. So, we need to find P(x>76) or P(x>-3).

With this value of $\\ z = -3$ , we can obtain this probability consulting the cumulative standard normal distribution, available in any Statistics book or on the internet.

Third: Determination of the probability P(x>76) or P(x>-3).

Most of the time, the values for the cumulative standard normal distribution are for positive values of z. Fortunately, since the normal distributions are symmetrical, we can find the probability of a negative z having into account that (for this case):

$\\ P(z>-3) = 1 - P(z>3) = P(z$

Then

Consulting a cumulative standard normal table, we have that the cumulative probability for a value below than three (3) standard deviations is:

$\\ P(z$

Thus, "the probability that a random selected student score is greater than 76" for this case (that is, $\\ \mu = 85$ and $\\ \sigma = 3$ ) is $\\ P(x>76) = P(z>-3) = P(z$ .