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1 year ago

Explain how to find the actual area of logo 2. Then provide the area of Logo 2.

Mathematics

1 answer:

marysya [2.9K]1 year ago

7 0

The actual area of logo 2 is the amount of space on it

The actual area of logo 2 is 31.25 square feet

<h3>How to determine the actual area of logo 2?</h3>

The scale area of logo 2 is calculated using the following equation

A = (1/2 * (1/2 * 3)) + (1/2 (1/2 * 2))

Evaluate the sum and the product

A = 1.25 square inches

The scale ratio is given as:

1 inches : 5 feet

Take the square of both sides

1 square inches : 25 square feet

So, the actual area is:

A = 1.25 * 25 square feet

A = 31.25 square feet

Hence, the actual area of logo 2 is 31.25 square feet

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In 1981, the Australian humpback whale population was 350 and has increased at a rate of 14% each year since then.)

dimulka [17.4K]

Answer:

In 1981, the Australian humpback whale population was 350

Po = Initial population = 350

rate of increase = 14% annually

P(t) = Po*(1.14)^t

P(t) = 350*(1.14)^t

Where

t = number of years that have passed since 1981

Year 2000

2000 - 1981 = 19 years

P(19) = 350*(1.14)^19

P(19) = 350*12.055

P(19) = 4219.49

P(19) ≈ 4219

Year 2018

2018 - 1981 = 37 years

P(37) = 350*(1.14)^37

P(37) = 350*127.4909

P(37) = 44621.84

P(37) ≈ 44622

There would be about 44622 humpback whales in the year 2018

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3 years ago

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aivan3 [116]

Answer:

Step-by-step explanation:

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2 years ago

A card is selected at random from an ordinary deck of 52 cards. What is the probability that it is either an ace or a spade

tamaranim1 [39]

Answer: 4/13

Step-by-step explanation:

Number of cards in a deck = 52

Number of spades = 13

Number of aces = 4

Number of ace of spade = 1

Probability of either a spade or an ace :

P(spade or ace) = P(spade U ace)

P(spade U ace) = p(spade) + p(ace) - p(spade n ace)

Probability = required outcome / Total possible outcomes

P(spade) = 13 / 52

P(ace) = 4 / 52

P(spade n ace) = 1 / 52

P(spade U ace) = p(spade) + p(ace) - p(spade n ace)

= 13/52 + 4/52 - 1/52

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2 years ago

A high school sewing club has 11 members. In how many different ways can they choose a president and vice president?

lorasvet [3.4K]

Since the order matters Person A and B can be chosen as AB (A pres B vice pres) or BA (B president A Vice president)

When the order matters we are talking about permutations. How many permutations of 11 taken 2 at a time are there.
It can be written like this 11nPr 2
It means start at 11 and multiply by the next lower number
11 x 10 = 110 ways they can choose.
Extension of the concept:
If there had been three positions President, Vice President, and Treasurer
then it would be 11nPr3 or 11 x 10 x 9
Once again the answer to your problem is 110.

6 0

3 years ago

Read 2 more answers

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Len [333]

Answer:

(1, 3) and (4,0)

Step-by-step explanation:

The quadratic function is $y=x^2-6x+8$