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nalin
1 day ago
14

Clayton places tiles that spell the word time in a bowl. He draws one tile from the bowl and then without replacing it, draws an

other tile. What is the sample space of the event representing the letters picked in the two draws? {T-I, T-M, T-E, I-M, I-E, M-E} {T-I, T-M, T-E, I-T, I-M, I-E, E-T, E-M, E-I, M-I, M-T, M-E} {T-I, T-T, T-E, I-T, I-I, I-E, E-T, E-M, E-E, M-T, M-M, M-E} {T-I, T-M, T-E, T-T, I-T, I-M, I-E, I-I, E-T, E-M, E-I, E-E, M-T, M-T, M-E, M-M}

Mathematics
2 answers:
babunello [8.4K]1 day ago
7 0
The scenario can be illustrated using a tree diagram as depicted below. The pairings of drawing the first and second tile form

TI IT MT ET
TM IM MI EI
TE IE ME EM

Certain combinations yield identical outcomes regardless of the order, such as TI and IT being equivalent. We can streamline the combinations to:

TI
TM
TE
MI
IE
ME
ET

AnnZ [9K]1 day ago
6 0

Answer:

The sample space for the selection of letters during two draws without replacement is:

{T-I, T-M, T-E, I-T, I-M, I-E, E-T, E-M, E-I, M-I, M-T, M-E}

Explanation in steps:

There are four options for drawing a tile first:

T I M E

Then:

  • If the first tile is T, the possible outcomes for the second draw are:

T-I T-M T-E

(T-T is not an option as there is no replacement)

  • Similarly, if I is drawn first, the outcomes for the second draw would be:

I-T I-M I-E

(I-I cannot happen due to no replacement)

  • When M is picked first, the possible outcomes corresponding to M are:

M-T M-I M-E

(M-M isn’t viable since there's no replacement)

  • If E is the first draw, the outcomes for E in the second draw would be:

E-T E-I E-M

(E-E cannot occur due to no replacement)

Thus, the complete set of outcomes is:

{T-I, T-M, T-E, I-T, I-M, I-E, E-T, E-M, E-I, M-I, M-T, M-E}

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A college football coach has decided to recruit only the heaviest 15% of high school football players. He knows that high school
Inessa [8989]

Response:

The coach should begin seeking players who weigh at least 269.55 pounds.

Step-by-step explanation:

We have these details from the question:

Average, μ = 225 pounds

Standard Deviation, σ = 43 pounds

The weights follow a bell curve, indicating a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We need to establish the value of x that corresponds to a probability of 0.15

P( X > x) = P( z > \displaystyle\frac{x - 225}{43})=0.15

= 1 -P( z \leq \displaystyle\frac{x - 225}{43})=0.15

=P( z \leq \displaystyle\frac{x - 225}{43})=0.85

Review from the standard normal z table gives us:

P(z < 1.036) = 0.85

\displaystyle\frac{x - 225}{43} = 1.036\\\\x = 269.548 \approx 269.55

Consequently, the coach should start recruiting players weighing at least 269.55 pounds.

3 0
1 month ago
How do I interpret this equation 5+1x10= is the answer 15 or 60?
Leona [9260]
To solve simple equations like this, it's essential to apply the order of operations defined by PEMDAS. This acronym represents the sequential operations needed for solving equations.
PEMDAS indicates Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction.
According to PEMDAS, multiplication takes precedence over addition, leading us to:
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6 0
1 month ago
Read 2 more answers
Midpoint of the segment between the points (−5,13) and (6,4)
Leona [9260]

Answer: (0.5,8.5)

Step-by-step explanation:

To find the Midpoint "M", we apply this formula:

M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

From the coordinates (-5,13) and (6,4), we determine that:

x_1=-5\\x_2=6\\\\y_1=13\\y_2=4

The final phase involves substituting the values into the formula.

Thus, the midpoint of the line segment connecting the points (-5,13) and (6,4) calculates to:

M=(\frac{-5+6}{2},\frac{13+4}{2})\\\\M=(\frac{1}{2},\frac{17}{2})\\\\M=(0.5,8.5)

6 0
3 days ago
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PIT_PIT [9101]

Answer:

Step-by-step explanation:

To find the combined function (f+g)(x), we proceed as follows:

(f+g)(x) = f(x) + g(x) = x^2 + 3x + 5 + x^2 + 2x = 2x^2 + 5x + 5

Therefore, the graph of (f+g)(x) depicts a parabola with the vertex located at (-5/4, 15/8).

I hope this is helpful.

5 0
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