1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Travka [436]
2 years ago
6

Find x if a =13 and C = 47 need help

Mathematics
1 answer:
dedylja [7]2 years ago
6 0

So you're using the Pythagorean theorem which is a^2 + b^2 = c^2... so if you dont have b you can manipulate the equation. That equation is c^2 - a^2= b^2, so you square 47 which is 2209 and then square 13 which is 169... you do 2209-169 and that equals 2040, then you do the square root of that which is 45.166359 that can be simplified to 45 or two decimal points 45.167. Hope that Helped! :)

You might be interested in
Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete
icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, 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 17 means that N = 17

4 are carrying nucleas weapons, which means that k = 4

9 are destroyed, which means that n = 9

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

P(X > 1) = 1 - P(X \leq 1)

In which

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

So

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

P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294

P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412

P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0
2 years ago
Proport
Gwar [14]
It takes 5 minutes per square foot (400/180=5) and so 5 minutes per square for 720 squares: 720x5=3,600. 3,600 minutes is equivalent to 60 hours so it takes him 60 hours.
4 0
3 years ago
Read 2 more answers
Which inequality represents the graph below
scoray [572]

Answer:

A

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
It says write in scientific notation but isn’t it already in scientific notation ???
Tems11 [23]
Oh..it sure yes hm..that’s odd
6 0
3 years ago
Read 2 more answers
Which is the product of 15 and 5/12 ?
zysi [14]

Answer:

6 1/4

Step-by-step explanation:

15 x 5/12 = 15 x 5/12 = 75/12

75/12= 6 3/12 = 6 1/4

4 0
2 years ago
Other questions:
  •   The population of a city is expected to increase by 7.5% next year.  If p represents the current population, which expression
    14·1 answer
  • Jake is skipping stones onto a pond. Each time he flings a stone in, he counts
    15·2 answers
  • Round to the nearest hundred thousand on a number 89,659
    6·1 answer
  • Lim as x approaches 2pi/3 from the right of csc x, solve by substituting csc with sin
    13·1 answer
  • X – y = 3 (line a)
    15·1 answer
  • Solve the quadratic equation m2+ 2m - 24 = 0 by factoring.
    13·1 answer
  • Solve the equation below <br><br> 2x-4=10<br><br><br> A. X=10<br> B. X=14<br> C.X=7<br> D.X=2
    10·2 answers
  • Find the L.C.M. of 16, 30 and 48​
    14·2 answers
  • 5x - 12 = 13, solve using algebra.
    12·2 answers
  • 10.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!