8/100 in decimal form

What is the domain of the function f(x)= 1/3x+2

frutty [35]

If it is 1/(3x+2), 3x+2≠0 =>x≠-2/3, the domain is all real number except -2/3
if it is 1/(3x)+2, 3x≠0, x≠0, the domain is all real number except 0

It is hard to tell from your equation what the demonstrator is.

3 0

4 years ago

Read 2 more answers

ADD together all the numbers greater than 0.7, in the following list . check your answer by adding the numbers together in a dif

sertanlavr [38]

Answer: 15.28

Step-by-step explanation: I just added them together

6 0

4 years ago

If x^2+y^2=1, what is the largest possible value of |x|+|y|?

Marianna [84]

If x² + y² = 1, then y = ±√(1 - x²).

Let f(x) = |x| + |±√(1 - x²)| = |x| + √(1 - x²).

If x < 0, we have |x| = -x ; otherwise, if x ≥ 0, then |x| = x.

• Case 1: suppose x < 0. Then

f(x) = -x + √(1 - x²)

f'(x) = -1 - x/√(1 - x²) = 0 → x = -1/√2 → y = ±1/√2

• Case 2: suppose x ≥ 0. Then

f(x) = x + √(1 - x²)

f'(x) = 1 - x/√(1 - x²) = 0 → x = 1/√2 → y = ±1/√2

In either case, |x| = |y| = 1/√2, so the maximum value of their sum is 2/√2 = √2.

6 0

3 years ago

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) $P(400$

$P(-1$

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:

$P(X$