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LenaWriter [7]
3 years ago
11

ms Klein surveyed some students to find where they wanted to go for. a field trip. she gave them four choices and recorded the r

esults in a table
Mathematics
1 answer:
Ivanshal [37]3 years ago
7 0
What is the question

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PLEASE HELP<br><br> Which line segment is a radius of K?<br> A) EH<br> B) FG<br> C) DK<br> D) KH
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Answer:1.5 i think

Step-by-step explanation:

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Please help thanks math
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Answer should be 63 because you multiply to find area
8 0
3 years ago
Find a fraction equivalent to 5/7 whose squared terms add up to 1184.
bogdanovich [222]

The system of equations of two unknowns is formulated and solved.

\large\displaystyle\text{$\begin{gathered}\sf \bf{ \left\{\begin{matrix} \ \ \ \dfrac{x}{y} = \dfrac{5}{7} \\ x^2+y^2 = 1184 \end{matrix}\right. \ \Longrightarrow \ x=\dfrac{5}{7}y  } \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf \bf{ \left (\dfrac{5}{7}y \right )^2+y^2=1184\ \Longrightarrow\ 25y^2+49y^2=58016 } \end{gathered}$}

                                                              \large\displaystyle\text{$\begin{gathered}\sf \bf{74y^{2}=58016} \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf \bf{ \ \  \ \ \ \ y^{2}=784 } \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf \bf{ \ \ \ \ \ y=\pm\sqrt{784}=\pm28  } \end{gathered}$}

                                                                  \large\displaystyle\text{$\begin{gathered}\sf \bf{ x=\dfrac{5}{7}(\pm 28)=\pm 20  } \end{gathered}$}

The fraction that satisfies the request is \bf{\dfrac{20}{28}} , since in \bf{\dfrac{-20}{-28}} the negative signs are canceled and the first fraction is obtained.

3 0
2 years ago
Given a standard deck of 52 cards, 3 cards are dealt without replacement. Using this situation, answer the questions below.&lt;b
kherson [118]
Given that <span>3 cards are dealt without replacement in a </span><span>standard deck of 52 cards.

Part A:

There are 4 queens in a standard deck of 52 card, thus the probability that the first card is a queen is given by 4 / 52 = 1 / 13.

Since, the first card is not replaced, thus there are 3 queens remaining and 51 ards remaining in total, thus the probability that the second card is a queen is given</span> by 3 / 51 = 1 / 17

Similarly the probability that the third card is a queen is given by 2 / 50 = 1 / 25.

Therefore, the probability that <span>all three cards are queens is given by

\frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} = \frac{1}{5525}



Part B:

Yes the probability of drawing a queen of heart is independent of the probability of drawing a queen of diamonds because they are separate cards and drawing one of the cards does not in any way affect the chance of drawing the other card.



Part C:

Given that the first card is a queen, then there are 3 queens remaining out of 51 cards remaining, thus the number of cards that are not queen is 51 - 3 = 48 cards.

Therefore, </span>if the first card is a queen, the probability that the second card will not be a queen is given by 48 / 51 = 16 / 17



Part D:

<span>Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining.

Therefore, </span>if two of the three cards are queens ,<span>the probability that you will be dealt three queens</span> is given by 2 / 50 = 1 / 25 = 0.04



Part E:

<span>Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining, thus the number of cards that are not queen is 50 - 2 = 48 cards.

Therefore, </span>if two of the three cards are queens ,the probability that the other card is not a queen is given by 48 / 50 = 24 / 25 = 0.96
8 0
3 years ago
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