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3 years ago

Regular triangular pyramid has 6 cm long base edge and slant height k=9 cm. Find the lateral area of the pyramid.

Mathematics

2 answers:

Yuri [45]3 years ago

4 0

Answer

81 cm^2

Step-by-step explanation:

lateral area : 3 triangles of base 6 cm and height 9 cm

Lateral area = 3 times 1/2 times 6 times 9

= 81 cm^2

cricket20 [7]3 years ago

3 0

The lateral area consists of 3 triangles of base 6 cm and height 9 cm

Lateral area = 3 * 1/2 * 6 * 9 = 81 cm^2 answer

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There are 175 baskets of fruit in the supermarket. 13 baskets of them are bananas. What percent of the baskets are not bananas?

alexira [117]

Based on the information given the percent of the baskets that are not bananas is: 92.57%.

<h3>Percent of the baskets that are not bananas </h3>

Using this formula

Percentage=(Number of baskets of fruit-Basket of bananas)/Number of baskets of fruit

Let plug in the formula

Percentage=(175-13)/175

Percentage=162/175×100

Percentage=92.57%

Inconclusion the percent of the baskets that are not bananas is: 92.57%.

Learn more about percentage here:brainly.com/question/24304697

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2 years ago

6x+4=22<br><br> What is the value of x?

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3 years ago

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Which of the following numbers is Irrational?

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3 years ago

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A certain firm has plants A, B, and C producing respectively 35%, 15%, and 50% of the total output. The probabilities of a non-d

Sliva [168]

Answer:

There is a 44.12% probability that the defective product came from C.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

-In your problem, we have:

P(A) is the probability of the customer receiving a defective product. For this probability, we have:

$P(A) = P_{1} + P_{2} + P_{3}$

In which $P_{1}$ is the probability that the defective product was chosen from plant A(we have to consider the probability of plant A being chosen). So:

$P_{1} = 0.35*0.25 = 0.0875$

$P_{2}$ is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

$P_{2} = 0.15*0.05 = 0.0075$

$P_{3}$ is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

$P_{3} = 0.50*0.15 = 0.075$

$P(A) = 0.0875 + 0.0075 + 0.075 = 0.17$

P(B) is the probability the product chosen being C, that is 50% = 0.5.

P(A/B) is the probability of the product being defective, knowing that the plant chosen was C. So P(A/B) = 0.15.

So, the probability that the defective piece came from C is:

$P = \frac{0.5*0.15}{0.17} = 0.4412$

There is a 44.12% probability that the defective product came from C.

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4 years ago

Select the correct answer.

lara [203]

First convert them to fractions (the answer choices are fractions):

a) 0.333... = 1/3
b) 0.555... = 5/9

Now add a and b:

1/3 (or 3/9 for a common denominator) + 5/9 = 8/9

That fraction is in simplest form, so leave it as it is.

Hope this helps!

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4 years ago