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3 years ago

What is the median of the given set of data? 1. 1. 5. 4. 3. 7. 2.9. 4, 8

Mathematics

1 answer:

san4es73 [151]3 years ago

6 0

Answer:

Step-by-step explanation:

Arrange in order

1,1,2,3,4,4,5,7,8,9

Cross off each number from each side little by little

1,2,3,4,4,5,7,8

2,3,4,4,5,7

3,4,4,5

4,4

There are two numbers left so we would find whats between them. But, since it is the same number, you just write the same one as the median.

median= 4

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PLEASE HELP. Prove that the value of the expression: b (36^5−6^9)(38^9−38^8) is divisible by 30 and 37.

pishuonlain [190]

Answer:

Step-by-step explanation:

First factor the second binomial

38^9 - 38^8 = 38^9 (38 - 1 ) = 38^9(37)

So there's your 37. This whole expression is divisible by 37

Now do the first binomial

36^5 - (36^4)*6

36^4(36 - 6)

36^4(30) and there's your thirty.

That first term is going to cause a bit of trouble showing that 36^4 * 6 = 6^9

6(36^4) = 6(6^2)^4 = 6^1 * 6^8 = 6^9

So this factors into 30*(36^4)*37*38^9

3 0

3 years ago

Your friend asks if you would like to play a game of chance that uses a deck of cards and costs $1 to play. They say that if you

gtnhenbr [62]

Answer:

Expected value = 40/26 = 1.54 approximately

The player expects to win on average about $1.54 per game.

The positive expected value means it's a good idea to play the game.

============================================================

Further Explanation:

Let's label the three scenarios like so

scenario A: selecting a black card
scenario B: selecting a red card that is less than 5
scenario C: selecting anything that doesn't fit with the previous scenarios

The probability of scenario A happening is 1/2 because half the cards are black. Or you can notice that there are 26 black cards (13 spade + 13 club) out of 52 total, so 26/52 = 1/2. The net pay off for scenario A is 2-1 = 1 dollar because we have to account for the price to play the game.

-----------------

Now onto scenario B.

The cards that are less than five are: {A, 2, 3, 4}. I'm considering aces to be smaller than 2. There are 2 sets of these values to account for the two red suits (hearts and diamonds), meaning there are 4*2 = 8 such cards out of 52 total. Then note that 8/52 = 2/13. The probability of winning $10 is 2/13. Though the net pay off here is 10-1 = 9 dollars to account for the cost to play the game.

So far the fractions we found for scenarios A and B were: 1/2 and 2/13

Let's get each fraction to the same denominator

1/2 = 13/26
2/13 = 4/26

Then add them up

13/26 + 4/26 = 17/26

Next, subtract the value from 1

1 - (17/26) = 26/26 - 17/26 = 9/26

The fraction 9/26 represents the chances of getting anything other than scenario A or scenario B. The net pay off here is -1 to indicate you lose one dollar.

-----------------------------------

Here's a table to organize everything so far

$\begin{array}{|c|c|c|}\cline{1-3}\text{Scenario} & \text{Probability} & \text{Net Payoff}\\ \cline{1-3}\text{A} & 1/2 & 1\\ \cline{1-3}\text{B} & 2/13 & 9\\ \cline{1-3}\text{C} & 9/26 & -1\\ \cline{1-3}\end{array}$

What we do from here is multiply each probability with the corresponding net payoff. I'll write the results in the fourth column as shown below

$\begin{array}{|c|c|c|c|}\cline{1-4}\text{Scenario} & \text{Probability} & \text{Net Payoff} & \text{Probability * Payoff}\\ \cline{1-4}\text{A} & 1/2 & 1 & 1/2\\ \cline{1-4}\text{B} & 2/13 & 9 & 18/13\\ \cline{1-4}\text{C} & 9/26 & -1 & -9/26\\ \cline{1-4}\end{array}$

Then we add up the results of that fourth column to compute the expected value.

(1/2) + (18/13) + (-9/26)

13/26 + 36/26 - 9/26

(13+36-9)/26

40/26

1.538 approximately

This value rounds to 1.54

The expected value for the player is 1.54 which means they expect to win, on average, about $1.54 per game.

Therefore, this game is tilted in favor of the player and it's a good decision to play the game.

If the expected value was negative, then the player would lose money on average and the game wouldn't be a good idea to play (though the card dealer would be happy).

Having an expected value of 0 would indicate a mathematically fair game, as no side gains money nor do they lose money on average.

7 0

2 years ago

8. Hassan made a vegetable salad with 2 3/8 pounds

amm1812

let's convert firstly the mixed fractions to improper fractions and then add up.

$\bf \stackrel{mixed}{2\frac{3}{8}}\implies \cfrac{2\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{19}{8}}~\hfill \stackrel{mixed}{1\frac{1}{4}}\implies \cfrac{1\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{5}{4}} \\\\\\ \stackrel{mixed}{2\frac{7}{8}}\implies \cfrac{2\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{23}{8}} \\\\[-0.35em] ~\dotfill$

$\bf \cfrac{19}{8}+\cfrac{5}{4}+\cfrac{23}{8}\implies \stackrel{\textit{using an LCD of 8}}{\implies \cfrac{(1)19+(2)5+(1)23}{8}}\implies \cfrac{19+10+23}{8} \\\\\\ \cfrac{52}{8}\implies \cfrac{13}{2}\implies 6\frac{1}{2}$

5 0

3 years ago

The equation 12x² + 4kx + 3 = 0 has real and equal roots, if

sleet_krkn [62]

Answer:

$k=\pm3$

Step-by-step explanation:

[...] if you can write the LHS as a perfect square, or if you can't spot a factorization of it right away, if and only if the discriminant $\Delta = b^2-4ac$ (or, if b is an even number, 1/4 of it) is zero.

Stare at it for a while. First term is $3(2x)^2$ , third term is $3(1)^2$ , we are missing a double product, but we can play with k. For the LHS to be $3(2x\pm1)^2 = 3(4x^2\pm4x+1) = 12x^2\pm12x+3$ you just need $4k= \pm12 \rightarrow k=\pm3$ .

Then number crunching it is. Set the discriminant to 0, solve for k

$\frac{\Delta}4 = 4k^2-12\cdot 3 =0 \rightarrow 4k^2=36 \\k^2 = 9 \rightarrow k=\pm3$

8 0

2 years ago

In the fall, you go to the local nursery and purchase a package of 10 rose plants. The clerk informs you that on average you can

LUCKY_DIMON [66]

Let success be that a plant dies in the winter and failure be that a plant survives the winter, then
p = 0.05, q = 1 - 0.05 = 0.95

a.) P(1 or more will not survive the winter) = 1 - P(all will survive the winter) = 1 - P(0) = 1 - 10C0 (0.05)^0 (0.95)^10 = 1 - 1(1)(0.5987) = 0.4013

b.) P(all will survive) = P(none will die) = P(0) = 0.5987

8 0

3 years ago

What is the median of the given set of data? 1. 1. 5. 4. 3. 7. 2.9. 4, 8​

What is the median of the given set of data? 1. 1. 5. 4. 3. 7. 2.9. 4, 8