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4 years ago

The definition of a circle uses the undefined term? A.Arc B.Line C.Plane D.Ray

Mathematics

2 answers:

MArishka [77]4 years ago

8 0

A. Arc
I hope this helps you

algol [13]4 years ago

3 0

A arc hope this helps.

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Approximately how many times greater is 2.3x10^-4 than 1.15x10^-8 ?

Minchanka [31]

Answer:

2.3 × $10^{-4}$ is 20,000 times greater than 1.15 × $10^{-8}$

Step-by-step explanation:

Given as :

The first number= x = 2.3 × $10^{-4}$

The second number= y = 1.15 × $10^{-8}$

Let The first number is z times greater than second number

i.e x = z × y

Or, z = $\dfrac{x}{y}$

Or, z = $\dfrac{2.3\times 10^{-4}}{1.15\times 10^{-8}}$

Or, z = 2 × $10^{4}$

∴ z = 20,000

So,The first number is 20,000 times greater than second number

Hence, 2.3 × $10^{-4}$ is 20,000 times greater than 1.15 × $10^{-8}$ . Answer

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4 years ago

A section of a railway has a slope of approximately 0.37. In this section, a vertical change of one unit corresponds to a horizo

Sedaia [141]

Slope $\frac{dy}{dx}=0.37$

We are given $dy=1$
So $\frac{1}{dx}=0.37$
⇒ $dx=\frac{1}{0.37}$ ≈2.70

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3 years ago

Wrong or Right ? PLS ANSWER ASAP !!!

kramer

Answer:

WRONG

Step-by-step explanation:

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3 years ago

Which value of c is a solution to this equation?

Burka [1]

C = 9 is the value of C

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3 years ago

Read 2 more answers

the numbers 1,2,3,4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacet. find t

Tresset [83]

Consider such events:

A - slip with number 3 is chosen;

B - the sum of numbers is 4.

You have to count $Pr(A|B).$

Use formula for conditional probability:

$Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}.$

1. The event $A\cap B$ consists in selecting two slips, first is 3 and second should be 1, because the sum is 4. The number of favorable outcomes is exactly 1 and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event $A\cap B$ is

$Pr(A\cap B)=\dfrac{1}{20}.$

2. The event $B$ consists in selecting two slips with the sum 4. The number of favorable outcomes is exactly 2 (1st slip 3 and 2nd slip 1 or 1st slip 1 and 2nd slip 3) and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event $B$ is

$Pr(B)=\dfrac{2}{20}=\dfrac{1}{10}.$

3. Then

$Pr(A|B)=\dfrac{\frac{1}{20} }{\frac{1}{10} }=\dfrac{1}{2}.$

Answer: $\dfrac{1}{2}.$

5 0

3 years ago