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2 years ago

A drawing has a scale of 3 in. : 2 ft. What is the scale factor of the drawing?

Mathematics

1 answer:

icang [17]2 years ago

3 0

The scale factor of the drawing as described is; 1.5in/ft

<h3>Scale factor of drawings</h3>

According to the question;

The given scale factor of the drawing is; 3 in: 2 ft.

In essence, 3 inches on the drawing board represents 2 ft of the object.

Hence, by finding the quotient of the units, we have;

3in/2ft = 1.5 in/ft

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Find the slope of the line that passes through points (4, -20) and (10, -16).

frozen [14]

The answer is c. I think

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3 years ago

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Consider randomly selecting a student at a large university. Let A be the event that the selected student has a Visa card, let B

Tems11 [23]

Answer:

0.75

Step-by-step explanation:

Given,

P(A) = 0.6, P(B) = 0.4, P(C) = 0.2,

P(A ∩ B) = 0.3, P(A ∩ C) = 0.12, P(B ∩ C) = 0.1 and P(A ∩ B ∩ C) = 0.07,

Where,

A = event that the selected student has a Visa card,

B = event that the selected student has a MasterCard,

C = event that the selected student has an American Express card,

We know that,

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

= 0.6 + 0.4 + 0.2 - 0.3 - 0.12 - 0.1 + 0.07

= 0.75

Hence, the probability that the selected student has at least one of the three types of cards is 0.75.

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2 years ago

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Answer:

1)B

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Step-by-step explanation:

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2 years ago

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A candy company claims that 20% of the candies in its bags are colored green. Steve buys 30 bags of 30 candies, randomly selects

Allisa [31]

Answer:

The probability of Steve agreeing with the company’s claim is 0.50502.

Step-by-step explanation:

Let X denote the number of green candies.

The probability of green candies is, p = 0.20.

Steve buys 30 bags of 30 candies, randomly selects one candy from each, and counts the number of green candies.

So, n = 30 candies are randomly selected.

All the candies are independent of each other.

The random variable X follows a binomial distribution with parameter n = 30 and p = 0.20.

It is provided that if there are 5, 6, or 7 green candies, Steve will conclude that the company’s claim is correct.

Compute the probability of 5, 6 and 7 green candies as follows:

$P(X=5)={30\choose 5}(0.20)^{5}(1-0.20)^{30-5}=0.17228\\\\P(X=6)={30\choose 6}(0.20)^{6}(1-0.20)^{30-6}=0.17946\\\\P(X=7)={30\choose 7}(0.20)^{7}(1-0.20)^{30-7}=0.15328$

Then the probability of Steve agreeing with the company’s claim is:

P (Accepting the claim) = P (X = 5) + P (X = 6) + P (X = 7)

= 0.17228 + 0.17946 + 0.15328

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Thus, the probability of Steve agreeing with the company’s claim is 0.50502.

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