Help me pls ?? 1. Suppose there is a raffle with numbered tickets 1

HELP ME PLZ AND IF YOU HELP ME 10 POINTS

mafiozo [28]

Answer:

3(c+4)

Step-by-step explanation:

It would usually be 3 x 4, but since there is a variable, it can`t be multiplied (unless you know what the value of the variable is). Thus the expression is 3(c+4).

6 0

3 years ago

Read 2 more answers

Help plz and thank have a nice day

dusya [7]

This is for the first page

Answer: answer to question 17 is r=32/7, answer to question 18 is m=-12, answer to question 19 is k=-9, answer to question 20 is p= No solution, answer to question 21 is x= infinite solutions, answer to question 22 is x= -4.

Step-by-step explanation:

This picture is for the second page

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:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.842$

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.

8 0

3 years ago

What is the lcm of 16, 24 and 40

ivann1987 [24]

The answer is 240

Explanation: The Least Common Multiple (LCM) is the smallest number that two or more numbers will divide into evenly. NOTE: to find LCM you first need to know how to find GCD.

First we will find LCM for first two numbers ( 16 and24 ).

Step 1: Find the GCD (Greatest Common Divisor ) of 16 and 24 which is 8.

Step 2: Multiply the numbers 16 and 24 together ( 16 * 24 = 384 )

Step 3: Divide the 384 with 8. (384/8 = 48)

So, the LCM of 16 and 24 is 48.

Now we will find the LCM of above result (48) and third number ( 40 ) using the same procedure.

The result of this part is 240

8 0

3 years ago

In the game of Pick-A-Ball, there are 10 colored balls: 2 red, 4 white, and 4 blue. The balls have been placed into a small buck

MaRussiya [10]

Answer:

The probability of selecting the color of ball that you just selected(a red ball) is $P(r) = \frac{2}{10}$

Step-by-step explanation:

From the question we are told that

The total number of colored balls is n = 10

The number of red balls is r = 2 red

The number of white ball is w = 4

The number of blue balls is b = 4

The number of balls to be selected is N = 2 balls

The cost of playing picking a ball is c = $10

The winnings is $k = \$ P(A)$

Here P(A) is the probability of obtaining the exact sequence of result and from the question it is P(A) = 0.10

Generally my price according to the question is

$k = \frac{0.10 }{10}$

=> $k = 0.01$

Let assume we selected a red ball

Generally the probability of selecting a red ball is mathematically represented as

$P(r) = \frac{2}{10}$

8 0

3 years ago

Help me pls ?? 1. Suppose there is a raffle with numbered tickets 1 - 150. One ticket is randomly