Mathematics: Acoin is tossed,and a four-section spinner is spun once.the tree diagram shows the possible outcomes for the two events. what is the probability of flipping heads on the coin and spinning 4 on the spinner?

teamsStep-by-step explanation:Let s do it by parts. Let s consider a fact: there has to be only one winner. After six games:So, Since it s a decreasing Geometric Sequence (we have to lead to one winner), and each match two teams play, the beaten one leaves the championship.a) Explicit Rule (Formula):That s to say the tournament started with 32 teamsb) Recursive RuleIf we have the first term of the Geometric Sequence we can recursively find the subsequent terms. The Recursive Rule is useful when...

Acoin is tossed,and a four-section spinner is

Ответ:

jalenjoseph7527

16.03.2022

Mathematics

I believe that the answer is 1/8. I hope this helps

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Ответ:

fanta47

16.03.2022

Mathematics

Probability of getting heads on the coin and spinning 4 on the spinner = $\frac{1}{8}$

Step-by-step explanation:

When a coin is tossed the probability to get head or tail = $\frac{\text{favorable outcome}}{\text{total outcomes}}$

= $\frac{1}{2}$

When a spinner, divided in four sections numbered from 1 to 4 is spun the probability to get 4 on the spinner = $\frac{\text{favorable outcome}}{\text{total outcomes}}$

= $\frac{1}{4}$

Now probability of flipping heads on the coin and spinning 4 on the spinner = Probability of getting heads on the coin × probability of spinning 4 on the spinner

= $\frac{1}{2}\times \frac{1}{4}$

= $\frac{1}{8}$

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Ответ:

samuelerlikhman

28.11.2021

Mathematics

teams

Step-by-step explanation:

Let's do it by parts.

Let's consider a fact: there has to be only one winner. After six games:

So, $a_{6}=1$

Since it's a decreasing Geometric Sequence (we have to lead to one winner), and each match two teams play, the beaten one leaves the championship.

$r=(\frac{1}{2})$

$a_{1}=?$

a) Explicit Rule (Formula):

$a_{6}=1,a_{1}=?r=\frac{1}{2}\\1=a_{1}\left ( \frac{1}{2} \right )^{6-1}\\1=a_{1}\left ( \frac{1}{2} \right )^{5}\\1=\frac{a_{1}}{32}\\a_{1}=32$

That's to say the tournament started with 32 teams

$\{32,16,8,4,2,1\}$

b) Recursive Rule

If we have the first term of the Geometric Sequence we can recursively find the subsequent terms. The Recursive Rule is useful when we have the prior term and we need the following one. Also, to derive the explicit rule.

$a_{1}=32\\a_{n}=\left ( \frac{1}{2} \right )a_{n+1}\\a_{2}=\left ( \frac{1}{2} \right )32=16\\a_{3}=\left ( \frac{1}{2} \right )16=8\\a_{4}=\left ( \frac{1}{2} \right )8=4\\a_{5}=\left ( \frac{1}{2} \right )4=2\\a_{6}=\left ( \frac{1}{2} \right )2=1\\$

$\{32,16,8,4,2,1\}$

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Acoin is tossed,and a four-section spinner is spun once.the tree diagram shows the possible outcomes for the two events. what is the probability of flipping heads on the coin and spinning 4 on the spinner?

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