I need help x-x please

What is the Domain and Range *DON'T FORGET PARENTHESES AND BRACKETS*

Alborosie

Answer:

D= {-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6}

R={-6,-5,-4,-3,-2,-1,1,2,5)

Step-by-step explanation:

5 0

3 years ago

Serggg [28]

Answer:

B. 252

Step-by-step explanation:

To find the answer, you have to use the formula to calculate the number of combinations as in this case, the order in which the objects are selected doesn't matter:

nCr=n!/r!(n-r)!

n= number of sample points: 10

r= number of sample points in each combination: 5

10C5= 10!/5!(10-5)!

10C5= 10!/5!*5!

10C5= 252

According to this, the answer is that you have 252 different ways to choose your school lunch.

5 0

4 years ago

Laura works at a coffee shop. Her boss has asked that she write down how many minutes each of the next 10 orders take to make. H

ASHA 777 [7]

Answer:

4

Step-by-step explanation:

6+2+4+2+2+7+5+6+3+3=40

40/10=4

Therefore the mean is 4

8 0

3 years ago

A positive integer from one to six is to be chosen by casting a die. Thus the elements c of the sample space C are 1, 2, 3, 4, 5

Alex_Xolod [135]

Answer with Step-by-step explanation:

We are given that six integers 1,2,3,4,5 and 6.

We are given that sample space

C={1,2,3,4,5,6}

Probability of each element= $\frac{1}{6}$

We have to find that $P(C_1),P(C_2),P(C_1\cap C_2) \;and\; P(C_1\cup C_2)$

Total number of elements=6

$C_1$ ={1,2,3,4}

Number of elements in $C_1$ =4

$P(E)=\frac{number\;of\;favorable \;cases}{Total;number \;of\;cases}$

Using the formula

$P(C_1)=\frac{4}{6}=\frac{2}{3}$

$C_2$ ={3,4,5,6}

Number of elements in $C_2$ =4

$P(C_2)=\frac{4}{6}=\frac{2}{3}$

$C_1\cap C_2$ ={3,4}

Number of elements in $(C_1\cap C_2)=2$

$P(C_1\cap C_2)=\frac{2}{6}=\frac{1}{3}$

$C_1\cup C_2=$ {1,2,3,4,5,6}

$P(C_1\cup C_2)=\frac{6}{6}=1$

4 0

3 years ago

To graduate this semester, you must pass Statistics 314; and you estimate that you have an 85% chance of passing. You need to pa

Nitella [24]

Answer:

Probability of graduating this semester is 0.7344

Step-by-step explanation:

Given the data in the question;

let A represent passing STAT-314

B represent passing at least in MATH-272 or MATH-444

M1 represent passing in MATH-272

M2 represent passing in MATH-444

C represent passing GERMAN-32

now

P( A ) = 0.85, P( C ) = 90, P( M1 ) = P( M2 ) = 0.8

P( B ) = P( pass at least one of either MATH-272 or MATH-444 ) = P( pass in MATH-272 but not MATH-444 ) + ( pass in MATH-444 but not in MATH 272) + P( pass both )

P( B ) = P( M1 ) × ( 1 - P( M2 ) ) + ( 1 - P( M1 ) ) × P( M2 ) + P( M1 ) × P( M2 )

we substitute

⇒ 0.8×0.2 + 0.2×0.8 + 0.8×0.8 = 0.16 + 0.16 + 0.64 = 0.96

∴ the probability of graduating this semester will be;

⇒ P( A ) × P( B ) × P( C )

we substitute

⇒ 0.85 × 0.96 × 90

⇒ 0.7344

Probability of graduating this semester is 0.7344

6 0

3 years ago