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Naddik [55]
3 years ago
11

Rationalise the denominator of 5√7 / √5

Mathematics
1 answer:
Tomtit [17]3 years ago
6 0
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u><u> </u></h2>

Rationalize the denominator of 5√7/ √5

<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u> </u></h2>

5√7 / √5

=> 5√7/√5×√5/√5

=> 5√7×5/(√5)^2

=> 5√35/5

=>√35

=> 5√7

<h2><u>S</u><u>o</u><u>,</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>r</u><u>e</u><u>q</u><u>u</u><u>i</u><u>r</u><u>e</u><u>d</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u> </u><u>i</u><u>s</u><u> </u><u>5</u><u>√</u><u>7</u><u> </u><u>.</u><u> </u></h2>
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The square root 5 lies between which two whole numbers
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SAT scores are normed so that, in any year, the mean of the verbal or math test should be 500 and the standard deviation 100. as
vovangra [49]

Answer:

a) P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)

P(Z>1.25)=1-P(Z

b) P(400

P(-1

P(-1

c) z=-0.842

And if we solve for a we got

a=500 -0.842*100=415.8

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.  

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".  

Part a

Let X the random variable that represent the SAT scores of a population, and for this case we know the distribution for X is given by:

X \sim N(500,100)  

Where \mu=500 and \sigma=100

We are interested on this probability

P(X>625)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)

And we can find this probability using the complement rule and with the normal standard table or excel:

P(Z>1.25)=1-P(Z

Part b

We are interested on this probability

P(400

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

P(400

And we can find this probability with this difference:

P(-1

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(-1

Part c

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.8   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(Z>-0.842)=0.8

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.842

And if we solve for a we got

a=500 -0.842*100=415.8

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.  

8 0
3 years ago
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