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const2013 [10]
3 years ago
6

Solve this equation 6(4.5y-12)=9

Mathematics
1 answer:
irinina [24]3 years ago
6 0
6(4.5y - 12) = 9

(4.5y - 12) = 9/6

4.5y - 12 = 1.5

4.5y = 1.5 + 12

4.5y = 13.5

y = 13.5/4.5

y = 3 
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Cho tam giác ABC vuông ở A có AB=12 cm, AC=5cm
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English pls coz I don’t understand
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2 years ago
Help please, stuck on this one!
allochka39001 [22]

Answer:

AFCE = 30

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Step-by-step explanation:

From the picture, you can see the figure can be decomposed into two identical large triangles with area 20, and 4 identical longer ones, that have to cover the remaining area 40, thus 10 each.

From that you can count the area.

Second puzzle is just trial and error.

4 0
2 years ago
if someone apply to 5 different colleges and 4 different majors how many different scenarios are there
svetoff [14.1K]

Answer:

He can apply for 4 different majors at each college giving

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4 0
2 years ago
Write an equation to represent the following: Three consecutive integers add up to 11 more than the smallest.
mr Goodwill [35]

Answer:


Step-by-step explanation:

First we have to assume the smallest integer.

Let,

the smallest integer is  x

So, the other two integers are-

x+1  and   x+2

Now,  

the sum of these 3 integers is greater than smallest integer by  11

So, we get

x+(x+1)+(x+2) =x+11

This is the required equation.

5 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.  

5 0
3 years ago
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