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erik [133]
3 years ago
12

Add the fractions and simplify the answer 1/2 +2 /3

Mathematics
2 answers:
Brilliant_brown [7]3 years ago
5 0
So first convert the denoenator (bottom n umber) to the sam enumber


1/2 and 2/3
the numbers are 2 and 3
find least comon multiple
easy way to convert is multiply each by bottom number of the other over the other exg
1/2 +2/3
1/2 times 3/3+2/3 times 2/2
3/6+4/6
(7)/6
7/6
1 and 1/6
IrinaVladis [17]3 years ago
4 0
What we are going to do is add \frac{1}{2} by \frac{2}{3}. The first thing we do is make common denominators for \frac{1}{2} and \frac{2}{3}:\frac{1*3}{2*3}= \frac{3}{6} \ and \  \frac{2*2}{3*2}= \frac{4}{6}. Now we add \frac{3}{6} by \frac{4}{6}: \frac{3}{6}+ \frac{4}{6}= \frac{7}{6}. Now we make \frac{7}{6} a mixed number:\frac{7}{6}=1 \frac{1}{6}. Therefore, your answer is \boxed{\boxed{1 \frac{1}{6}}}. Hoped I helped. :)
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Answer:

z = \frac{418.7-465.6}{\frac{53.6}{\sqrt{7}}}= -2.340

Now we can calculate the p value with the following probability

p_v =2*P(z

If we use the commonly significance level of 0.05 we see that the p value is lower than these values we can conclude that the true mean is different from the value of 465.6 at 5% of significance

Step-by-step explanation:

Data provided

\bar X=418.7 represent the mean score on the standardized memory test

\sigma=53.6 represent the population standard deviation

n=7 sample size    

\mu_o =465.6 represent the value that we want to verify

z would represent the statistic

p_v represent the p value

System of hypothesis

We are interested to check if the memory performance for elderly individuals differs from that of the general population (mean different from 465.6), the system of hypothesis are then:

Null hypothesis:\mu = 465.6    

Alternative hypothesis:\mu \neq 465.6    

Since we know the population deviation we can use the z statistic from the z test for the true mean:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}} (1)    

Replacing the info given we got:

z = \frac{418.7-465.6}{\frac{53.6}{\sqrt{7}}}= -2.340

Now we can calculate the p value with the following probability

p_v =2*P(z

If we use the commonly significance level of 0.05 we see that the p value is lower than these values we can conclude that the true mean is different from the value of 465.6 at 5% of significance

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You can set it up as an equation:

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I'm not sure how this really relates to the distributive property though.
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