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3 years ago

What is the measure, in degrees, of the largest angle in a right triangle?

Mathematics

1 answer:

anzhelika [568]3 years ago

7 0

Im guessing the largest side of the triangle, i mean cmon. its the largest.

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Fill in the space <br> -1•6=(_)•(-1)<br> Commutative or associative

iren [92.7K]

Answer:

Step-by-step explanation:

I am pretty sure this is your correct answer because you just switch the order

6 0

3 years ago

nventing is a difficult way to make money. Only 5% of new patents earn a substantial profit. A certain city has just had30 indep

Arlecino [84]

Answer:

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

And we can find the individual probabilities using the probability mass function and we got:

$P(X=0) = (30C0) (0.05)^0 (1-0.05)^{30-0} =0.2146$

$P(X=1) = (30C1) (0.05)^1 (1-0.05)^{30-1} = 0.3389$

And replacing we got:

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

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=30, p=0.05)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Solution to the problem

For this case we want this probability:

$P(X \geq 2)$

And we can use the complement rule and we got:

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

And we can find the individual probabilities using the probability mass function and we got:

$P(X=0) = (30C0) (0.05)^0 (1-0.05)^{30-0} =0.2146$

$P(X=1) = (30C1) (0.05)^1 (1-0.05)^{30-1} = 0.3389$

And replacing we got:

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

3 0

3 years ago

Help asap please !!!

Igoryamba

Answer:

ROTATION

Step-by-step explanation:

5 0

3 years ago

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Convert 4/6 into a fraction with a denominator of 24

deff fn [24]

The correct answer would be 16/24. You would need to multiply both the denominator and numerator by 4.

3 0

4 years ago

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What number must you add to complete the square?<br> x^2 + 2x = -1

Andru [333]

$\large\begin{array}{l} \textsf{You must add 1 to complete the square.}\\\\ \textsf{There is a very common special product: (square of a sum)}\\\\ \mathsf{(a+b)^2=a^2+2ab+b^2}\\\\\\ \textsf{Take the given equation:}\\\\ \mathsf{x^2+2x=-1\qquad(i)}\\\\\\ \textsf{The first term is a square (x squared)}\\\\ \textsf{The second term is a product of 2, x and 1.} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{We need to find a proper value for b, so that if we add it to}\\\textsf{the left side of the equation, we get a perfect square.}\\\\ \textsf{So,}\\\\ \begin{array}{cccccccccc} \mathsf{a^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{a}&\!\!\!\cdot\!\!\!&\mathsf{b}&\!\!\!+\!\!\!&\mathsf{b^2}&=\mathsf{(a+b)^2}\\\\ \mathsf{\downarrow}&&&\!\!\!\!\!&\downarrow&\!\!\!\!\!&\downarrow&&\downarrow\\\\ \mathsf{x^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{x}&\!\!\!\cdot\!\!\!&\mathsf{1}&\!\!\!+\!\!\!&\mathsf{1^2}&=\mathsf{(x+1)^2} \end{array}\\\\\\ \textsf{Just by comparing the expressions above, we can conclude that}\\\textsf{b must be equal to 1, so we get a perfect square:}\\\\ \mathsf{(x+1)^2=x^2+2x+1} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{Back to (i), adding 1 to both sides:}\\\\ \mathsf{x^2+2x+1=-1+1}\\\\ \boxed{\begin{array}{c} \mathsf{(x+1)^2=0} \end{array}} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{If you want to solve this equation for x, you will find}\\\\ \mathsf{x+1=0}\\\\ \mathsf{x=-1}\quad\longleftarrow\quad\textsf{(solution)} \end{array}$

If you're having problems understanding the answer, try seeing it through your browser: brainly.com/question/2112248

$\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best regards! :-)} \end{array}$

6 0

3 years ago

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