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vekshin1
3 years ago
7

The sum of all positive integers less than 400 which are divisible by 5. Can someone give me the steps to do this?

Mathematics
1 answer:
IceJOKER [234]3 years ago
3 0

Answer:

Sum of all positive integers less than 400 and divisible by 5 is 15,800.

Step-by-step explanation:

TO FIND :

The sum of all positive integers less than 400 which are divisible by 5.

The set of positive integers I^+  =  1,2,3, 4,5,6,7,.........,,

Now, the number should be divisible by 5.

SO, the desired set of positive integers =  { 5,10,15,20,.......}

Again the numbers are LESS than 400.

So, the desired set of positive integers =  { 5,10,15,20,....... 385,390, 395}

Here, First term a  = 5, common difference d  = 4 and last term an = 395

a_n = a+ (n-1) d\\\implies 395 = 5 + (n-1) 5\\\implies 78 = n - 1 \implies n =  79

⇒There are total 79 terms in the series.

So, SUM OF 79 TERMS  =

S_n = \frac{n}{2} (a + a_n) \\\implis S_{79} = \frac{79}{2} (5 + 395)  = 15,800\\\implies S_{79} = 15,800

Hence, The sum of all positive integers less than 400 which are divisible by 5 is 15,800.

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jarptica [38.1K]
-\frac{1}{9} =-0.1111111111...
8 0
3 years ago
It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de
scoray [572]

Answer:

a) P(X

P(z

b) P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)

P(z>-0.583)=1-P(Z

c) P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)

P(z>2.33)=1-P(Z

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) z=1.28

And if we solve for a we got

a=72000 +1.28*12000=87360

So the value of height that separates the bottom 90% of data from the top 10% is 87360.  

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 life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

X \sim N(72000,12000)  

Where \mu=72000 and \sigma=12000

We are interested on this probability

P(X

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

And we can find this probability using excel or the normal standard table and we got:

P(z

Part b

P(X>65000)

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>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)

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

P(z>-0.583)=1-P(Z

Part c

P(X>100000)

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>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)

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

P(z>2.33)=1-P(Z

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

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

P(X>a)=0.1   (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 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

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=1.28

And if we solve for a we got

a=72000 +1.28*12000=87360

So the value of height that separates the bottom 90% of data from the top 10% is 87360.  

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The equation will be;

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Now multiply equation 2 with (2)

The equation will be;

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Now add equation 3 and equation 4

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<span>6a  + 10b = -4</span>

<span>------------------------------</span>

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2a -3 = -1

2a = -1+3

2a = 2

a=1

Thus the solution is (a,b) = (1,-1)

<span>
</span>

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The value of 2 in 2783 (2000) is 10 times the value of two in 7283 (200).

Hope this helped!

Nate

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a 12-ft-long ladder is leaning against a building. the ladder makes a 45 degree angle with the building. how far up the building
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12 cos45degree=8.486 ft
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