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2 years ago

A rectangular retaining wall has an area of

Mathematics

1 answer:

Veronika [31]2 years ago

8 0

Answer:

The height of wall is 3 feet and length of wall is 5 feet

Step-by-step explanation:

Given as :

The Area of retaining wall = 15 square feet

Let The width of wall = W feet

The Height of wall = ( W - 2 ) feet

Now, Area of Rectangle = Height width

Or , 15 ft² = ( W - 2 ) ft × W ft

Or, 15 = W² - 2 W

Or, W² - 2 W - 15 = 0

Or, W² - 5 W + 3 W - 15 = 0

Or, W ( W - 5 ) + 3 ( W - 5 ) = 0

So , ( W - 5 ) ( W + 3 ) = 0

So, W = 5 , - 3

∴ The width of wall = W = 5 feet

And The Height of wall = ( W - 2 ) = 5 - 2 = 3 feet

Hence The height of wall is 3 feet and length of wall is 5 feet. Answer

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B. 16 feet

Step-by-step explanation:

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1 year ago

Need help with this please!!

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3 years ago

A marathon is 26.2 miles long sue wants to run a marathon in 4 hours what is the average number of miles she must run in an hour

harkovskaia [24]

Answer:

6.55 miles in one hour

Step-by-step explanation:

26.2 divided by 4 = 6.55

that would give you the miles in one hour.

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2 years ago

K - 2( k - ( 2- k ))-2

Naily [24]

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3 years ago

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r

kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that $p = 0.15$

Assume you obtain a random sample of 9 individuals from this population:

This means that $n = 9$

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

$P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005$

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0

2 years ago