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2 years ago

Help me pleaseee!! I’ve been stuck on it for a while:(

Mathematics

2 answers:

dimulka [17.4K]2 years ago

5 0

Answer:

Step-by-step explanation:

I did not do the math becals I could not find my pen but based on outher math like this I have a strong feeling it's a if not the maby d

Norma-Jean [14]2 years ago

3 0

Answer:

its the 3rd one :)

Step-by-step explanation:

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You are choosing numbers from the set S ={A, B, C, D, E, F, G, H}. Let x = the number of ways to choose 3 letters from S without

postnew [5]

Given set S = <span>{A, B, C, D, E, F, G, H}
There are 8 elements in set S and we are to choose 3 letters at random, the number of ways to choose such is x. It is simply similar to choosing 5 letters at random, which is also equal to x. Since order doesn't matter, n! / (n-m)! where n = 8 and m = 3, which is 336 ways. </span>

6 0

2 years ago

Sandra wants to purchase necklaces that cost $4.99. She already purchased 3 bracelets for $8.97. Which of the following function

Volgvan

The answer is A

Y= 4.99x + 8.97

6 0

3 years ago

What is the value of the expression below?

Ghella [55]

Answer:

A. 12

Step-by-step explanation:

use pemdas

4^4=16

20+8=28

28-16=12

4 0

2 years ago

Read 2 more answers

Please help Evaluate this equation!!

lozanna [386]

Answer:

1.)-8

2.)-5

3.)4

Step-by-step explanation:

So, the function f(x)=3x-5

The questions say "If you were to substitute these numbers what answer would you get?"

Understanding this,

First question is with x=-1

f(-1)=3(-1)-5

f(-1)=-3-5=-8

_____________________________________

f(0)=3(0)-5

f(0)=0-5=-5

_____________________________________

f(3)=3(3)-5

f(3)=9-5=4

6 0

2 years ago

Suppose two $20 bills, three $10 bills, one $5 bill, and seven $1 bills are placed in a bag. If you were to pull a bill at rando

Korolek [52]

Answer:

Assuming that the $1 bill was pulled at random, then the expected value of the amount chosen is $\frac{7}{13}$ .

Step-by-step explanation:

From the given question, the bag contains;

$1 bill = 7

$5 bill = 1

$10 bill = 3

$20 bill = 2

Total number of bills in the bag = 13

Pulling a bill at random, the bills would have an expected value as follows:

For $1 bill, the expected value = $\frac{7}{13}$

For $5 bill, expected value = $\frac{1}{13}$

For $10 bill, expected value = $\frac{3}{13}$

For $20 bill, the expected value = $\frac{2}{13}$

Assuming that the $1 bill was pulled at random, then the expected value of the amount chosen is $\frac{7}{13}$ .

6 0

3 years ago