All categories

3 years ago

An equilateral triangle has one length of 3 meters. What are the lengths of the other two sides?

Mathematics

1 answer:

topjm [15]3 years ago

5 0

Answer:

the other sides are 3 meters as well

Step-by-step explanation: Since it is an equilateral triangle, all the sides will be the same. An Isosceles triangle has two of the sides the same and one different. A scalene triangle has all of the sides different. So, the rest of the sides are 3 meters.

You might be interested in

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper

Korvikt [17]

Answer:

90.67% probability that John finds less than 7 golden sheets of paper

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it contains a golden sheet of paper, or it does not. The probability of a container containing a golden sheet of paper is independent of other containers. So 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.

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper containers.

This means that $p = 0.3$

14 of these containers of paper.

This means that $n = 14$

What is the probability that John finds less than 7 golden sheets of paper?

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{14,0}.(0.3)^{0}.(0.7)^{14} = 0.0068$

$P(X = 1) = C_{14,1}.(0.3)^{1}.(0.7)^{13} = 0.0407$

$P(X = 2) = C_{14,2}.(0.3)^{2}.(0.7)^{12} = 0.1134$

$P(X = 3) = C_{14,3}.(0.3)^{3}.(0.7)^{11} = 0.1943$

$P(X = 4) = C_{14,4}.(0.3)^{4}.(0.7)^{10} = 0.2290$

$P(X = 5) = C_{14,5}.(0.3)^{5}.(0.7)^{9} = 0.1963$

$P(X = 6) = C_{14,6}.(0.3)^{6}.(0.7)^{8} = 0.1262$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0068 + 0.0407 + 0.1134 + 0.1943 + 0.2290 + 0.1963 + 0.1262 = 0.9067$

90.67% probability that John finds less than 7 golden sheets of paper

7 0

3 years ago

Enter the equation of the circle described below.<br><br> Center (-1, 4), radius = √3

lorasvet [3.4K]

EQUATION OF A CIRCLE: (x-h)² + (y-k)² =r²

= (x+1)² + (y-4)² =3

3 0

4 years ago

How solve this <br> x+y=5<br> X-y=7

Misha Larkins [42]

$elimination\ methode\\\\\underline{+\left\{\begin{array}{ccc}x+y=5\\x-y=7\end{array}\right}\ \ \ \ |add\ both\ sides\ of\ the\ equations\\.\ \ \ \ \ \ 2x=12\ \ \ \ \ |divide\ both\ sides\ by\ 2\\.\ \ \ \ \ \ \ \boxed{x=6}\\\\subtitute\ the\ value\ of\ x\ to\ the\ first\ equation:\\\\6+y=5\ \ \ \ |subtract\ 6\ from\ both\ sides\\\boxed{y=-1}\\\\Answer:\boxed{\left\{\begin{array}{ccc}x=6\\y=-1\end{array}\right}$