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Vitek1552 [10]
4 years ago
9

In a probability experiment Eric flip a coin 36 times. The coin landed on head 24 times. What is the ratio of heads to tails in

this experiment?
A.3/2
B.1/2
C.2/3
D.2/1
Mathematics
2 answers:
ValentinkaMS [17]4 years ago
8 0

Answer:

2/3 is the correct answer so C. Hope this helps you. :)

Step-by-step explanation:

noname [10]4 years ago
3 0

Answer:

2/3 is the correct answer so C. Hope this helps you. :)

Step-by-step explanation:

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Lin missed class and Tyler is helping her use the table to approximate the angle measures that have the ratios listed. Tyler say
Lapatulllka [165]

Answer:

d is the correct answer

Step-by-step explanation:

bc i did thid

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3 years ago
I really need help with this. I'd really appreciate anyone's help.
vekshin1

Answer:

I think it's G

Step-by-step explanation:

Since a=8 root 3=8×3=24cm

Yh so I multiplied the 24cm by 12cm

12cm×24cm=288cm squared

and the square root of 288cm is 96 root 3 cm squared

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I need help with this math question
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Third option. The letter has neither rotate or symmetry
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CalculatorAC
Anarel [89]

Answer:

18 degrees heres youre answer!

Step-by-step explanation:

If Angle 1 is x, and Angle 2 is 9x (because one angle is 9 times as large as the other), then x + 9x = 180. Add like terms: 10x = 180. solve for x by dividing both sides by 10. Therefore, x = 18 degrees = smaller angle.

6 0
2 years ago
Prove c(x) with the supremum norm is a metric space showing the supremum nom is a metric. 1.
zhuklara [117]

Answer:

Step-by-step explanation:

The best way to solve this problem is to prove that any norm defines a metric, and then proving that the supremum norm is, in fact, a norm.

<em>Let us prove that any norm defines a metric.</em>

Assume that \|\cdot\| is a norm in a linear space E. Define

d(x,y)=\|x-y\|

and let us prove that d is a metric.

The first to axioms of metric are trivial:

  • \|x-y\|\geq 0 for all x,y \in E. Moreover, 0=d(x,y)=\|x-y\| and \|x-y\|=0 if and only if x-y=0.
  • d(x,y)=\|x-y\| = \|y-x\| =d(y,x).

As usual, the triangular inequality is less simple, but not so hard in this case:

d(x,y) = \|x-y\| = \|x-z+z-y\| \leq \|x-z\| + \|z-y\| = d(x,z)+d(z,y)

and this holds for every x,y,z\in E. Recall that from the definition of norm we already have a triangular inequality: \| x+y\| \leq \|x\| + \|y\|.

Now, we are in conditions to prove that the supremum norm, is a norm.

<em>The supremum norm is a norm on </em>\mathcal{C}(X).

The supremum norm is defined as \|f\|_\infty=\sup_{x\in[a,b]}|f(x)|, where f is a continuous function over the set X. Next, we are going to prove the three axioms.

(N1): \displaystyle f\equiv 0\Rightarrow\sup_{x\in[a,b]}|f(x)|=0.

                   \displaystyle\sup_{x\in X} |f(x)|=0\Rightarrow\forall x\in E\;\;|f(x)|\leq 0 \Rightarrow f(x)=0\Rightarrow f\equiv 0

(N2): \displaystyle\|\lambda f\|_\infty=\sup_{x\in X}|\lambda f(x)|=\sup_{x\in E}|\lambda| |f(x)|=|\lambda|\sup_{x\in X}|f(x)|=|\lambda|\|f(x)\|_\infty

(N3): \displaystyle \|f+g\|_\infty=\sup_{x\in X}|f(x)+g(x)|

                   \forall x\in X\;\;|f(x)+g(x)|\leq|f(x)|+|g(x)| from here we get

               \displaystyle\sup_{x\in X}|f(x)+g(x)|\leq\sup_{x\in X}(|f(x)|+|g(x)|) \leq \sup_{x\in X}|f(x)|+\sup_{x\in X}|g(x)|=\|f\|_\infty+\|g\|_\infty

8 0
3 years ago
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