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1 year ago

Triangle ABC , m∠A=45° and c = 16, find b.

Mathematics

1 answer:

zmey [24]1 year ago

5 0

Using the cosine function:

$\begin{gathered} \cos (\theta)=\frac{_{\text{ }}adjacent}{_{\text{ }}hypotenuse} \\ so\colon \\ \cos (45)=\frac{b}{16} \end{gathered}$

Solve for b:

$\begin{gathered} b=16\cdot\cos (45) \\ b=8\sqrt[]{2} \end{gathered}$

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What is 8 divided by 827

Dima020 [189]

The answe is 103.375, because when you divide 827 by 8, you get the answer of 103.375. To be sure, you may also type it in a calculator, and you'll noticed you'll get the same answer.

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3 years ago

HELP ASAP George plans to cover his circular pool for the upcoming winter season. The pool has a diameter of 20 feet and extends

enyata [817]

The pool has a diameter 20 ft so: r = 10 ft.
The pool cover extents 12 inches beyond the edge of the pool.
12 inches = 1 foot
Therefore, the radius of the pool cover is : r = 10 + 1 = 11 ft.
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3 years ago

Naval intelligence reports that 99 enemy vessels in a fleet of 1818 are carrying nuclear weapons. If 99 vessels are randomly tar

oee [108]

Answer:

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

Step-by-step explanation:

The vessels are destroyed and then not replaced, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

In this question:

Fleet of 18 means that $N = 18$

9 are carrying nuclear weapons, which means that $k = 9$

9 are destroyed, which means that $n = 9$

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,18,9,9) = \frac{C_{9,0}*C_{9,9}}{C_{18,9}} = 0.000021$

$P(X = 1) = h(1,18,9,9) = \frac{C_{9,1}*C_{9,8}}{C_{18,9}} = 0.001666$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.000021 + 0.001666 = 0.001687$

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

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3 years ago

Find the sum 1/2 + 3/8

Stels [109]

Answer: 7/8

Step-by-step explanation:

The easiest way to do this is to find a common denominator, so first we find out 2 times how much equals 8? 2x4 = 8. So with 8 being our common denominator, we have to multiple 4 to every number in the fraction 1/2. So 1x4 = (4/8) = 2x4. Then you just add the top numbers, so 4 + 3 = 7 and keep the denominator. 7/8.

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3 years ago

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erastovalidia [21]

I think the answer is C.

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3 years ago

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