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kiruha [24]
3 years ago
13

A group of 4 friends is playing cards. The deck has 70 cards. To start the game, the dealer makes a pile of 15 cards in the cent

er. Then she deals the remaining cards one at a time to each player until all the cards are gone. What is the greatest number of cards any player will have after all the cards are dealt?
Mathematics
1 answer:
ale4655 [162]3 years ago
4 0

Answer:

14 cards

Step-by-step explanation:

70 - 15 = 55

55/4 = 13.75

Therefore, the highest number of cards someone is able to get is 14.

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Answer:

divide the figure into two parts.

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area of triangle = 1/2×3×3

area pf triangle = 4.5 units²

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area of square = 3²

area of square = 9unit²

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Mary is rolling a standard six-sided die and spinning on a spinner with the number 1-8.
Readme [11.4K]

Answer:

a) Dice and spinner are different objects, with no relation between themselves, and thus her dice rolls and spinning outcomes are independent.

b) \frac{1}{48} probability of her rolling a 4 and spinning an 8.

c) \frac{1}{8} probability of her rolling an even number and spinning a number less than 3.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

A) Are her dice rolls and spinning outcomes independent or dependent?

Dice and spinner are different objects, with no relation between themselves, and thus her dice rolls and spinning outcomes are independent.

B) What is the probability of her rolling a 4 and spinning an 8?

Since the events are independent, we find each separate probability and multiply them.

Probability of rolling a 4:

One side out of 6 on the dice. So

P(A) = \frac{1}{6}

Probability of spinning an 8:

One side out of 8 on the spinner. So

P(B) = \frac{1}{8}

Probability of rolling a 4 and spinning an 8:

P(A \cap B) = P(A)P(B) = \frac{1}{6} \times \frac{1}{8} = \frac{1}{48}

\frac{1}{48} probability of her rolling a 4 and spinning an 8.

C) What is the probability of her rolling an even number and spinning a number less than 3?

Since the events are independent, we find each separate probability and multiply them.

Probability of rolling an even number:

2, 4 or 6(3 sides out of 6). So

P(A) = \frac{3}{6} = \frac{1}{2}

Probability of spinning a number less than 3:

Two numbers(1 or 2) out of 8 on the spinner. So

P(B) = \frac{2}{8} = \frac{1}{4}

Probability of rolling an even number and spinning a number less than 3:

P(A \cap B) = P(A)P(B) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}

\frac{1}{8} probability of her rolling an even number and spinning a number less than 3.

8 0
2 years ago
Drag each equation to the correct location on the table. Classify each function as linear or nonlinear.
drek231 [11]

Answer:

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Because I did the work

8 0
3 years ago
Find the minimum value of the function f(x)=x^2+5x-6
mash [69]
<h3><u>Explanation</u></h3>
  • Method 1 (Formula)

\begin{cases}h =  -  \frac{b}{2a}  \\ k =  \frac{4ac -  {b}^{2} }{2a}  \end{cases}

The vertex of Parabola is the maximum/minimum point depending on the value of a.

  • Find Vertex

<u>h-value</u>

h =  -  \frac{5}{2(1)}  \\ h =  -  \frac{5}{2}

<u>k-value</u>

k =  \frac{4(1)( - 6) -  {(5)}^{2} }{4(1)}  \\ k =  \frac{ - 24 - 25}{4}  \\ k =  \frac{ - 49}{4}  \\ k =  -  \frac{49}{4}

The minimum value is the value of k. Therefore the minimum value is - 49/4 at x = -5/2.

  • Method 2 (Derivative)

This is Calculus method. We simply differentiate the function then substitute y' = 0.

  • Differentiate Function

f'(x) = 2 {x}^{2 - 1}  +  {5x}^{1 - 1}  - 0 \\ f'(x) = 2x + 5

Substitute f'(x) = 0

0 = 2x + 5 \\  - 5 = 2x \\   -  \frac{5}{2}  = x

Substitute x = -5/2 in the original equation.

f(x) =  {( -  \frac{5}{2} )}^{2}  + 5( -   \frac{5}{2} ) - 6 \\ f(x) =  \frac{25}{4}  -  \frac{25}{2}  - 6 \\ f(x) =  \frac{25}{4}  -  \frac{50}{4}  -  \frac{24}{4}  \\ f(x) =  \frac{25}{4}  -  \frac{74}{4}  \\ f(x) = -   \frac{49}{4}

<h3><u>Answer</u><u /></h3>

<u>\sf{the  \:  \: minimum  \:  \: value \:  \:  is   \:  \: -  \frac{49}{4}  \:  \: at \:  \:  x =  -  \frac{5}{2} }</u>

4 0
3 years ago
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