What is the area of the square shown below? a9ft .b9ft2. c12ft. d.12ft2

If two quadrilateral are congruent, which congruence statement is correct?

postnew [5]

If the two quadrilateral are congruent, then ABCD = PQRS is the correct congruence statement.

4 0

3 years ago

Due to random variations in the operation of an automatic coffee machine, not every cup is filled with the same amount of coffe

DaniilM [7]

Answer:

465

because it is the wright answer

5 0

3 years ago

write the given equation into vertex form and sketch the graph. Determine the domain and range f(x)=3x^2-6+2 plss help

LuckyWell [14K]

Answer: the range would thus be [-3, 1/3].)

Step-by-step explanation:

1. To sketch the function f(x) = (2/3)x - 3, we first need to find two points that we can later join to sketch the line, for example the x- and y-intercepts.

a) The x-intercept occurs when f(x) = 0, so if f(x) = 0, then:

f(x) = (2/3)x - 3

0 = (2/3)x - 3

3 = (2/3)x (Add three to both sides)

3*(3/2) = x (Multiply both sides by 3/2)

9/2 = x

We have now found the x-intercept at (9/2, 0)

b) The y-intercept occurs when x = 0, so:

f(x) = (2/3)x - 3

f(0) = (2/3)*0 - 3

f(0) = -3

Now we know that the y-intercept is at (0, -3)

c) All that's left is to sketch the graph axes and label them, plot the two points, join them together using a ruler and label their coordinates.

2. The Domain is the range of x-values for which the function exists, and the Range is the range of y-values for which the function exists.

Since there haven't been any constraints specified, we can say that both the Domain and Range are (-∞, ∞), since the graph continues forever both along the x- and y-axis.

(Note that this isn't always the case and would change if, for example, the question stipulated that there was a domain of [0, 5] and you had to find the range. Then, you would calculate the value of y at each end of the domain (if x = 0, y = -3 and if x = 5, y = 1/3) - in my example, the range would thus be [-3, 1/3].)

4 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:

$P(X$