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3 years ago

What is the area of a triangle with a base of 15 1/4 inches and a height of 10 inches?

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

3 0

Area of the triangle is 76.25 in²

Step-by-step explanation:

Step 1: Find the area of the triangle with base = 15 1/4 in and height = 10 in

Area of the triangle = 1/2 × base × height

= 1/2 × 61/4 × 10 = 76.25 in²

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2 years ago

Suppose box A contains 4 red and 5 blue poker chips and box B contains 6 red and 3 blue poker chips. Then a poker chip is chosen

sergejj [24]

Answer:

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Coin chosen from box B is red.

Event B: Blue poker chip transferred.

Probability of choosing a red coin:

7/10 of 4/9(red coin from box A)

6/10 of 5/9(blue coin from box A). So

$P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444$

Blue chip transferred, red coin chosen:

6/10 of 5/9. So

$P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333$

What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172$

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

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2 years ago

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Yakvenalex [24]

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3 years ago

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Nadya [2.5K]

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3 years ago

Prove the sum of two rational numbers is rational where a, b, c, and d are integers and b and d cannot be zero. Fill in the miss

mario62 [17]

Answer:

We conclude that the sum of two rational numbers is rational.

Hence, the fraction will be a rational number. i.e.

$\frac{ad+cb}{bd}$ ∵ b≠0, d≠0, so bd≠0

Step-by-step explanation:

Let a, b, c, and d are integers.

Let a/b and c/d are two rational numbers and b≠0, d≠0

Proving that the sum of two rational numbers is rational.

$\frac{a}{b}+\frac{c}{d}$

As the least common multiplier of b, d: bd

Adjusting fractions based on the LCM

$\frac{a}{b}+\frac{c}{d}=\frac{ad}{bd}+\frac{cb}{db}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{ad+cb}{bd}$

As b≠0, d≠0, so bd≠0

Therefore, we conclude that the sum of two rational numbers is rational.

Hence, the fraction will be a rational number. i.e.

$\frac{ad+cb}{bd}$ ∵ b≠0, d≠0, so bd≠0

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3 years ago