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Vsevolod [243]
4 years ago
10

Hudson is performing an experiment by tossing a paper cup into the air and recording how it lands. He determines that there are

three possible outcomes for how the cup will land: on its top, on its bottom, or on its side. He is curious if the three outcomes are equally likely. The table below shows the results of Hudson's experiment.
Outcome Frequency
Top 4
Bottom 1
Side 25

a. What is the approximate probability that on the next toss, the paper cup will land on its top? On its bottom? On its side? Express each answer as a percentage.
b. From the results of Hudson's experiment, do the three outcomes appear to be equally likely? Explain your reasoning using complete sentences.
Mathematics
1 answer:
Debora [2.8K]4 years ago
3 0
It has landed on its side 25/30 times, on the top 4/30 times 25/30=5/6,1/30=0.2/6 and 4/30=0.8/6.
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Help me please<br>help.me​
levacccp [35]

Answer:

The answers of the questions are given below :

  • a) = 4096
  • b) = 1.25
  • 3) = m²
  • 4) = r⁴s³
  • 5) = a⁸/b¹²

Step-by-step explanation:

\large{\tt{\underline{\underline{\red{QUESTION}}}}}

\begin{gathered}\footnotesize\boxed{\begin{array}{c|c|c}\bf\underline{Given}&\bf\underline{Solution}&\bf{\underline{Simple\: Form}}\\\\\rule{60pt}{0.5pt} &\rule{70pt}{0.5pt}& \rule{70pt}{0.5pt}\\\\ 1.\: {4}^{6} & & \\\\ 2.\: \bigg(\dfrac{2^6}{5^3} \bigg)& &\\\\  3. \: \Big({m}^{\frac{2}{3}}\Big)\bull \Big({m}^{\frac{4}{3}}\Big) & &\\\\4. \:  \big({r}^{12} {s}^{9}\big)^{ - \frac{1}{3}} &&\\\\ 5.\bigg(\dfrac{a^4}{a^6}\bigg)^{2}& &\end{array}}\end{gathered}

\begin{gathered}\end{gathered}

\large{\tt{\underline{\underline{\red{SOLUTION}}}}}

Question. 1

>> 4⁶

\begin{gathered}\qquad{= 4 \times 4 \times 4 \times 4 \times 4 \times 4} \\  \qquad{= 16 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 64 \times 4 \times 4 \times 4} \\ \qquad{= 256\times 4 \times 4} \\ \qquad{= 1024  \times 4} \\ \qquad{= 4096} \end{gathered}

  • Hence, the answer is 4096.

\begin{gathered}\end{gathered}

Question. 2

>> (2⁶/5³)^-⅓

\begin{gathered} \qquad\implies{\bigg(\frac{2^6}{5^3}\bigg)^{ - \frac{1}{3}}}\\  \\ \qquad\implies{\bigg(\frac{64}{125}\bigg)^{ - \frac{1}{3}}}\\  \\\qquad\implies{\bigg( \frac{1}{\frac{64}{125}}\bigg)^{ \frac{1}{3}}} \\  \\ \qquad\implies{\bigg( 1 \times  \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\  \\ \qquad\implies{\bigg( \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\  \\\qquad\implies{\bigg( \sqrt[3]{ \frac{125}{64}}\bigg)}  \\  \\ \qquad\implies{\bigg( \dfrac{5}{4} \bigg)} \\  \\ \qquad\implies{\Big( 1.25\Big)}\end{gathered}

  • Hence, the answer is 1.25.

\begin{gathered}\end{gathered}

Question. 3

>> (m^2/3)•(m^4/3)

\begin{gathered} \qquad{=  \Big({m}^{\frac{2}{3}}\Big) \bull \Big({m}^{ \frac{4}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{\frac{2}{3} +  \frac{4}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{\frac{2 + 4}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{\frac{6}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{2}\Big)}\end{gathered}

  • Hence, the answer is m².

\begin{gathered}\end{gathered}

Question. 4

>> (r¹² s⁹)^⅓

\begin{gathered} \qquad\implies{\Big( {r}^{12} \: {s}^{9}\Big)^{\frac{1}{3}}}\\\\ \qquad\implies{\Big({r}^{\frac{12}{3} } \: {s}^{\frac{9}{3}}\Big)}  \\  \\ \qquad\implies{\Big({r}^{\cancel{\frac{12}{3}}} \: {s}^{\cancel{\frac{9}{3}}}\Big)}  \\  \\ \qquad\implies{\Big({r}^{4} \: {s}^{3}\Big)} \end{gathered}

  • Hence, the answer is r⁴s³.

\begin{gathered}\end{gathered}

Question. 5

>> (a⁴/b⁶)^2

\begin{gathered} \qquad{ =  \Big(\frac{a^4}{b^6}\Big)^{2}} \\ \\  \qquad{ =  \Big(\frac{a^{4 \times 2}}{b^{6 \times 2}}\Big)} \\ \\ \qquad{ =  \Big(\frac{a^{8}}{b^{12}}\Big)} \end{gathered}

  • Hence, the answer is a⁸/b¹².

\underline{\rule{220pt}{3pt}}

4 0
2 years ago
Write 2.72 as a mixed number and as a improper fraction
saveliy_v [14]

Answer:

your answer to this is 68/25

4 0
3 years ago
Dawn took two pies to a school bake. She cut Pie A into 8 slices and Pie B into 6 slices. She sold two slices of Pie A and three
evablogger [386]
2 of the 8 slices of pie A were sold (2/8)
3 of the 6 slices of pie B were sold (3/6)

Both of these fractions can be simplified

2/8 --> 1/4
3/6 --> 1/2

In order to add these two fractions, they must have a common denominator. This can be accomplished by multiplying the second fraction by 2

1/2 --> 2/4

Now we can add the two

1/4 + 2/4 = 3/4

Therefore 3/4 of the pie has been sold. In order to find out how much remains, we must subtract the amount that has been sold from the original amount of pie (2)

2 - 3/4 = 1 1/4 = 1.25
4 0
3 years ago
Jason's part-time job pays him $105 a week. If he has already saved $325, what is the minimum number of weeks he needs to work i
Art [367]

Jason's part-time job pays him $105 a week

He got an amount of 105 dollars per week

Thhe cost price of the Dit bike = $900

The saving amount he had = $325

So, The amount he require more is the difference in the cost price of the bike to the saving amount

Require amount = 900-325

Require Amount = $575

Since he get the $105 per week

We need to find the number of weeks to get an amount of $575

So, divide the total required amount by the amount in one week he get

\begin{gathered} \text{ Number of w}eek\text{ =}\frac{Total\text{ Amount}}{Amount\text{ he get p}er\text{ week}} \\ \text{Number of w}eek\text{ =}\frac{575}{105} \\ \text{Number of w}eeks\text{ =}5.47 \\ \text{Number of we}eks\text{ }\approx6 \end{gathered}

So, he neeed to wokr for 6 weeks to pay for the dirt bike

Answer : 6 weeks

8 0
1 year ago
Mr. Smith wants to order some pizza and cheese pizza cost $11 and pepperoni pizza cost $12.50 if you bought 13 pieces total and
Triss [41]

Answer: Mr Smith bought 7 pepperoni pizza

Step-by-step explanation:

Step 1

cost of cheese pizza  =$11

pepperoni pizza cost =$12.50

let number cheese pizza be represented as c

and number of pepperoni pizza be represented as p

such that the total number of pizza  bought which equals 13 can be represented as

c+ p= 13

and total cost of both pizza  can be expressed as

11c+ 12.50p =153.50

Step 2 Solving

 c+ p= 13

11c+ 12.50p =153.50

By elimination method, Multiply equation 1 by 11 and subtract the new equation from equation 2

11c+ 12.50p =153.50

11c+ 11p=143.

1.5p =10.5

p= 10.5/ 1.5 =7

c+ p= 13

c= 13-p =13-7=6

Therefore Mr Smith bought 6 cheese pizza and 7 pepperoni pizza

6 0
3 years ago
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