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3 years ago

What is the greatest common factor of 42a5b3, 35a3b4, and 42ab4?

1 answer:

4 0

For this case we have that by definition, the Greatest Common Factor (GCF) is the largest factor that 2 numbers have in common.

So, we have the following expressions:

$42a ^ 5b ^ 3\\35a ^ 3b ^ 4\\42ab ^ 4$

We look for the positive integers that divide to 35 and 42 without leaving residue:

42: 1,2,3,6,7,14,21

35: 1,5,7

Thus, the GCF of 42 and 35 is 7

Then, the GCF of the three expressions is:

$7ab ^ 3$

Answer:

$7ab ^ 3$

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Anyone can help me with math for solving for a side in the right triangles for khan academy

Elden [556K]

Answer:

7.45

Step-by-step explanation:

We are given a right triangle with one side length known and an angle know so we can use some trigonometry.

You need to find the hypotenuse so you are going to use either sine or cosine.

Since the adjacent from the angle is given (SOH-CAH-TOA) we can use cosine so,

cos(20) = 7/h

=>h = 7/cos(20)

=> 7.44924

=> 7.45

6 0

3 years ago

Use calculus to find the largest possible area for a rectangular field that can be enclosed with a fence that is 400 meters long

Setler79 [48]

the largest rectangular area is actually a square.

so the side would be 400/4 = 100 feet

area of a square is S^2

100^2 = 10,000 square feet

6 0

3 years ago

Which graph shows the solution set for 2 x + 3 greater-than negative 9?

sp2606 [1]

I''m pretty sure it's A

6 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:

$P(X$