All categories

3 years ago

The cost in dollars of a class party is 59 + 13n, where n is the number of people attending. What is the cost of 44 people?

Mathematics

1 answer:

SpyIntel [72]3 years ago

6 0

59 + 13n = 59 + 13(44) = 59 + 572 = 631 The cost of 44 people is $631.

You might be interested in

Mimi has 73 trading cards. Tanner has t more trading cards than Mimi. Write an expression that shows how many trading cards Tann

____ [38]

Hey! If Tanner has t more, just add to to 73, so the expression would be t+73.

<span>Hope that this helped! If it did, please give me a good rating and mark me brainliest for my hard work. Thanks!</span>

4 0

3 years ago

Read 2 more answers

Analyze the pre-image ABCD. What are the vertices of the final image if T–1, –2 ◦ ry = x is applied to figure ABCD? A'' B''(3, 2

Katen [24]

No one will understand if you don't add the picture but the answer from Edgenuity is

A: (3,0)

B: (3,2)

C: (1,1)

D: (1,-1)

Your Welcome

3 0

3 years ago

Read 2 more answers

a wildlife manager determines that the function y=200x1.07* represents the deer population at a state park x years after the pop

Kobotan [32]

Answer:

Table:

0,200

4, 262

8,344

12, 450

Population after 12 years is 450 deer

Initial population is 200 deer.

Step-by-step explanation:

You take the x in the table and plug it in as an exponent so you will need a calculator or multiple the 1.07 by that many times so 1.07 12 times multipled by itself. and the initial is 0 years so anything to the exponent of 0 is 1 so that show i got 200 for the x=0

8 0

3 years ago

In a standard deck of cards there 52 cards of which 26 are black and 26 are red. Additionally, there are 4 suits, hearts, clubs,

stich3 [128]

Using probability concepts, it is found that P(S and D) = 0.1275.

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

A probability is the <u>number of desired outcomes divided by the number of desired outcomes</u>.
In a standard deck, there are 52 cards.
Of those, 13 are spades, and 13 are diamond.

The probability of selecting a spade with the first card is 13/52. Then, there is a 13/51 probability of selecting a diamond with the second. The same is valid for diamond then space, which means that the probability is multiplied by 2. Thus, the desired probability is:

$P(S \cap D) = 2 \times \frac{13}{52} \times \frac{13}{51} = \frac{2\times 13 \times 13}{52 \times 51} = 0.1275$

Thus, P(S and D) = 0.1275.

A similar problem is given at brainly.com/question/12873219

7 0

2 years ago

Which one was it? I’ve been trying for a while

Reika [66]

The answer is b (I think not really sure)

6 0

3 years ago

Read 2 more answers