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jasenka [17]
3 years ago
15

Find the number that makes the ratio equivalent to 3:2. :24

Mathematics
2 answers:
frutty [35]3 years ago
5 0

Answer:

36:24

Step-by-step explanation:

24 divided by 2 which is 12 then 12x3 which is 36

I might be wrong but i think this is the answer.

Olin [163]3 years ago
4 0

Answer:  36 : 24

Step-by-step explanation:

3 / 2 = 12 * 3 / 12 * 2

= 36 / 24

= 36 : 24

Hope it helps !!

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How much is 1 and 3/4ths times 2
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To find out what 1 and 3/4 times 2 is, we need to find out a way to make it much more simple and then answer it.
A quick way to answer this in a few steps is to:
1.) convert the fraction into decimal.
To do this, we need to divide 100 by our denominator (the number on the bottom of a fraction). Let's try it.
100 / 4 = 25.
For every 1/4, it adds 25.
Since we have 3/4, multiply 3 by 25.
You get 75, but it is not whole.
2.) Put your quotient into decimal form by putting it as 0.## (if we divided 100 by a smaller number).
Your number for 3/4 is 0.75.
Since we have a whole number (1), put it in the single-digit place.
(#.00)
1.75 is your fraction into decimal form.
Now, we need to multiply 1.75 by 2.00.
2.00 x 1.75 = 3.5
Your product is 3.5.
I hope this helps!
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.<br><br> what is the area of the pentagon shown below?
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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 <em>parameters</em>, the <em>population mean</em> \\ \mu and the <em>population standard deviation</em> \\ \sigma.

To determine probabilities for the normal distribution, we can use <em>the standard normal distribution</em>, whose parameters' values are \\ \mu = 0 and \\ \sigma = 1. However, we need to "transform" the raw score, in this case <em>x</em> = 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 <em>z</em>. 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 <em>three</em> (3) <em>standard deviations</em> <em>below</em> 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 <em>the cumulative standard normal distribution, </em>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 <em>cumulative standard normal distribution</em> are for positive values of z. Fortunately, since the normal distributions are <em>symmetrical</em>, 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 <em>cumulative standard normal table</em>, 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.

As a conclusion, more than 99.865% of the values of this distribution are above (greater than) x = 76.

<em>We can see below a graph showing this probability.</em>

As a complement note, we can also say that:

\\ P(z3)

\\ P(z3)

Which is the case for the probability below z = -3 [P(z<-3)], a very low probability (and a very small area at the left of the distribution).

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