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3 years ago

Most everyday situations involving chance and likelihood are examples of ______.

Mathematics

2 answers:

Tpy6a [65]3 years ago

6 0

Most everyday situations involving chance and likelihood are examples of probability.

aleksklad [387]3 years ago

4 0

I believe the answer is probability

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Determine whether (18 + 35) × 4 = 18 + 35 ×4 is true or false. explain

fomenos

The first one is 212
and the second one is 158
so yes it is false they do not equal

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3 years ago

I need help figuring this problem out. <br> -x÷1/2=?

Lesechka [4]

-x divided by 1/2 = -2x

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3 years ago

My friend sets out walking at a speed of $3$ miles per hour. I set out behind her $5$ minutes later at $4$ miles per hour. When

r-ruslan [8.4K]

Answer: The answer is 0.25 miles.

Step-by-step explanation: Given that my friend started walking at a speed of 3 miles per hour and I started 5 minutes after him at a speed of 4 miles per hour. We need to find the number of miles my friend travelled when I started walking.

Since I started 5 minutes after than my friend, so we are to find the distance travelled by my friend in 5 minutes.

In 1 hour, i.e., 60 minutes, distance travelled by my friend = 3 miles.

In 1 minute, distance travelled by my friend is

$\dfrac{3}{60}=\dfrac{1}{20}~\textup{miles}.$

Therefore, distance travelled by my friend in 5 minutes will be

$\dfrac{1}{20}\times 5=\dfrac{1}{4}=0.25~\textup{miles}.$

Thus, the answer is 0.25 miles.

3 0

3 years ago

Read 2 more answers

PLEASE HELP!

Lina20 [59]

Answer:

Step-by-step explanation:

Each 90° clockwise turn takes a point (x,y) and transforms it to (y,-x). For a 180° turn you would do this process twice in a row; for a 270° turn, three times in a row.

4 0

3 years ago

Read 2 more answers

Atmospheric pressure decreases by about 11.8% for every 1000 meters you climb. The pressure at sea level is 1013 millibars (a un

erica [24]

At sea level, the pressure is 1013, that's when the altitude is at 0, sea level, let's see

$\bf \textit{Periodic Exponential Decay}\\\\
A=I(1 - r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &1013\\
I=\textit{initial amount}\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &0\\
p=period\to &1000
\end{cases}
\\\\\\
1013=I(1-0.118)^{\frac{0}{1000}}\implies 1013=I\cdot 1\implies 1013=I$

so, the inital amount is 1013, when t = 0,

$\bf \textit{Periodic Exponential Decay}\\\\
A=I(1- r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &1013\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &t\\
p=period\to &1000
\end{cases}
\\\\\\
A=1013(1-0.118)^{\frac{t}{1000}}\implies A=1013(0.882)^{\frac{t}{1000}}$

now, to check the atmospheric pressure at 4000, simply set t = 4000, to get A.

7 0

3 years ago