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krek1111 [17]
4 years ago
12

PLS HELP IM IN A TIMED TEST AND I DONT KNOW WHAT TO DO

Mathematics
1 answer:
fiasKO [112]4 years ago
5 0

Answer:

Hope you learned your lesson pay attention in class

Step-by-step explanation:

have faith don't cheat on a test

You might be interested in
I need help PLEASE.
bonufazy [111]
Notes: 
The notation ">=" without quotes means "greater than or equal to"
The upper case "U" means "set union"
Instead of using the intersection symbol, I will use a lower case 'n'

-------------------------------

Problem 1

A = {x | x < 1} which is the set of x values smaller than 1
B = {x | x >= 5} is the set of x values that are equal to 5 or larger
A U B = set of values that are from set A OR they are from set B (or both)
A U B = {x | x < 1 or x >= 5}
we simply connect the two inequalities mentioned with an "or" 

note: how there is no overlap between the two regions. The "U" means "set union" which is like a sort of glue to tie the two sets together with an "or". 

Answer: {x| x < 1 or x >= 5}
-------------------------------
Problem 2

A = {x | x < 1}
C = {x | x = 5} which is the set of one value only: 5 (x cannot equal any other value)
A U C = {x| x < 1 or x = 5}

So if a number is in set A U C, then this number is either less than 1, OR it is equal to 5

Answer: {x|x < 1 or x = 5}
-------------------------------
Problem 3

B U C = {x | x >= 5} because set C already has the "or equal to" part in there. 

Set C is a subset of set B. If an item is in set C, then it is also in set B.

Answer: {x| x >= 5}
-------------------------------
Problem 4

Again recall that I'm using an 'n' to indicate "set intersection" instead of the upside down "U" symbol

A n B is the set of items that are in BOTH sets A and B at the same time. From problem 1, I mentioned there's a gap. There is no x value that is both less than 1 AND greater than or equal to 5. So this means that

A n B = empty set

which we use the "O" with a slash through it. This is a special symbol to indicate "empty set"

Another way to write "empty set" is to use curly braces with nothing inside like so { } 

Answer: The "O" with a slash through it
-------------------------------
Problem 5

B n C is the same as {x| x = 5}

Why? Because if an item is in B n C, then it has to be in BOTH set B and set C at the same time. The only way this happens is if x = 5. If x is any other value, then it won't be in set C

Answer: {x| x = 5}
4 0
3 years ago
Kater has 24 T-shirts.Kater has 8 fewer pairs of shoes than pairs of pants. If the number of T-shirts Kater has is double the nu
olga_2 [115]
T=number of shirt
p=pants
s=shoes

t=24

8 fewer shoes than pants means s=p-8

t shirts are double pants, means t=2p


so

t=24
s=p-8
t=2p

solve

we can see
24=t=2p
24=2p
divide both sides by 2
12=p

sub into other equation
s=p-8
s=12-8
s=4



4 pairs  of shoes
6 0
3 years ago
If dab = cba, d = 130 and c = 4x - 14 x = ?
Dafna11 [192]

Answer:

38

Step-by-step explanation:

By CPCTC, angles C and D are congruent.

130 = 4x - 14

144 = 4x

x = 38

8 0
2 years ago
Read 2 more answers
You intend to draw a random sample in order to test a hypothesis about an unknown population mean. You will use a hypothesis tes
kompoz [17]

Answer:

  1. Determine the null and alternative hypotheses.
  2. Select a level of significance.
  3. Draw a random sample, and calculate the sample statistic.
  4. Decide whether to reject or fail to reject the null hypothesis.

Step-by-step explanation:

The steps that involve testing a hypothesis can be sorted like this:

First, we have to determine the null and alternative hypothesis. This is related directly to the claim we are trying to prove.

Second, we select a level of significance. This depends on how conservative we want to be with the conclusions. We have to do it before evaluating the test statistic, although it is not involved in its calculations.

Third, we draw a random sample, and calculate the sample statistic. This is the core of the hypothesis test, were we use the information of a sample to infer a characteristic of the population.

Fourth and last, with the results of the test statistic, we decide whether to reject or fail to reject the null hypothesis. This leads to the conclusions about the claims.

5 0
3 years ago
Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z
kompoz [17]

a.

f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}

is a proper joint density function if, over its support, f is non-negative and the integral of f is 1. The first condition is easily met as long as C\ge0. To meet the second condition, we require

\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}

b. Find the marginal joint density of X and Y by integrating the joint density with respect to z:

f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz

\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}

Then

\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy

\approx\boxed{0.12886}

c. This probability can be found by simply integrating the joint density:

\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz

\approx\boxed{0.012262}

7 0
3 years ago
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