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Aneli [31]
2 years ago
14

The average score on a biology test was 72.1 (1 repeats). write the average score using a fraction.

Mathematics
1 answer:
Yuri [45]2 years ago
6 0
72 \frac{1}{9}  1÷9 = 111....   2÷9 = 222.... and so on.

The proof is a little more difficult.

Think of all those repeating ones as a variable - let's call it n

so n= .111111......... repeating

How can we get a single one of those ones to jump across the decimal and be on the left side.   We can multiply all of those ones by 10.

10n (ten times the original number) = 1.1111111  (ones still go on forever)

Now here is the interesting part.  Let's take all the repeating ones in the first number we made away from the second number.

10n = 1. 1111111......
<u>-  n  =  .  1111111....
</u>9n   = 1    (all of the repeaters are gone and only the one we moved to the left
                   of the decimal is left)

Now let's divide by 9 to get n by itself
<u>9n</u>   = <u>1
</u>9        9

And voila!   n = 1/9

So to repeat 72.111... written as a fraction is 72\frac{1}{9}


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Hitung jarak di antara titik P(7, 1) dengan titik Q(7, 8) ​
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7 units

Step-by-step explanation:

Distance between 2 points

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

Thus, distance PQ

=\sqrt{(8 - 1)^{2}  + (7 - 7)^{2} }  \\ = \sqrt{7^{2} }  \\  = 7

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The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev
Nataliya [291]

Answer:

(a) Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = 0.15716

(b) Probability that a standard normal random variable will be between .3 and 3.2 = 0.3814

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with;

    Mean, \mu = 30.05 inches        and    Standard deviation, \sigma = 0.2 inches

Let X = A sheet selected at random from the population

Here, the standard normal formula is ;

                  Z = \frac{X - \mu}{\sigma} ~ N(0,1)

(a) <em>The Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = P(30.25 < X < 30.65) </em>

P(30.25 < X < 30.65) = P(X < 30.65) - P(X <= 30.25)

P(X < 30.65) = P(\frac{X - \mu}{\sigma} < \frac{30.65 - 30.05}{0.2} ) = P(Z < 3) = 1 - P(Z >= 3) = 1 - 0.001425

                                                                                                = 0.9985

P(X <= 30.25) = P( \frac{X - \mu}{\sigma} <= \frac{30.25 - 30.05}{0.2} ) = P(Z <= 1) = 0.84134

Therefore, P(30.25 < X < 30.65) = 0.9985 - 0.84134 = 0.15716 .

(b)<em> Let Y = Standard Normal Variable is given by N(0,1) </em>

<em> Which means mean of Y = 0 and standard deviation of Y = 1</em>

So, Probability that a standard normal random variable will be between 0.3 and 3.2 = P(0.3 < Y < 3.2) = P(Y < 3.2) - P(Y <= 0.3)

 P(Y < 3.2) = P(\frac{Y - \mu}{\sigma} < \frac{3.2 - 0}{1} ) = P(Z < 3.2) = 1 - P(Z >= 3.2) = 1 - 0.000688

                                                                                           = 0.99931

 P(Y <= 0.3) = P(\frac{Y - \mu}{\sigma} <= \frac{0.3 - 0}{1} ) = P(Z <= 0.3) = 0.61791

Therefore, P(0.3 < Y < 3.2) = 0.99931 - 0.61791 = 0.3814 .

 

3 0
2 years ago
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