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3 years ago

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In Sunday's Super Bowl, Tampa Bay ended up winning over Kansas City. With the win, Tampa Bay's Quarterback Tom Brady now has 7 C

hampionships out of 10 tries. What is Tom Brady's overall winning percentage %? Hint 7 out 10 is what percentage?

2 answers:

maria [59]3 years ago

6 0

Answer: 70% Brady is the goay

Step-by-step explanation:

Pavlova-9 [17]3 years ago

5 0

Answer:

KANAS CITY CHIEFS ARE THE BEST FRICK TOM BRADY HE IS OLD HE IS 43 YEARS OLD LIKE DUDE STOP UR TO OLD FRICK THE TAMPA BAY BUCS

Step-by-step explanation:

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you are making curtains by alternating strips of solid colored fabric and patterned fabric. the solid colored fabric costs 99¢ p

weeeeeb [17]

Hello,

to find the total cost (c) of the colored strips we have to multiply the number of strips (n) by the cost of each color strip (99¢), that as an algebraic function is:

$\boxed{c= n*99}$ [¢]

Now, if we use 3 solid color strips, that means that we use 4 patterned strips, so:

$Cost=3*0.99+4*1.25 \\ \boxed{Cost= 7.97}$

Answer: c=n*99 and Cost per curtain= $7.97

4 0

3 years ago

7 - 9x - 15 + 12x HELPPPP

Gekata [30.6K]

Simplify. Combine like terms

12x - 9x = 3x

7 - 15 = -8

3x - 8 is your answer

hope this helps

5 0

3 years ago

Read 2 more answers

What is 485,783 rounded to its greatest place?

natima [27]

Answer:

486

Step-by-step explanation:

if you mean 485.782 with a decimal then it would be 486

7 0

3 years ago

X^2+6x+8÷x^2-16 simplify

Misha Larkins [42]

Answer:

The answer is in the file below, I wasn't able to make it in text form.

6 0

2 years ago

A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d

cestrela7 [59]

Answer:

(10) Person B

(11) Person B

(12) $P(5\ or\ 6) = 60\%$

(13) Person B

Step-by-step explanation:

Given

Person A $\to$ 5 coins (records the outcome of Heads)

Person $\to$ Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

$Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}$

$n(Head) = 32$

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

$Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}$

$n(Dice) = 30$

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Head)}$

$Pr(5) = \frac{1}{32}$

$Pr(5) = 0.03125$

From person B list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Dice)}$

$Pr(5) = \frac{8}{30}$

$Pr(5) = 0.267$

<em>From the above calculations: </em> $0.267 > 0.03125$ <em> Hence, person B is more likely to get 5</em>

Question 11: Person with Higher median

For person A

$Median = \frac{n(Head) + 1}{2}th$

$Median = \frac{32 + 1}{2}th$

$Median = \frac{33}{2}th$

$Median = 16.5th$

This means that the median is the mean of the 16th and the 17th item

So,

$Median = \frac{3+2}{2}$

$Median = \frac{5}{2}$

$Median = 2.5$

For person B

$Median = \frac{n(Dice) + 1}{2}th$

$Median = \frac{30 + 1}{2}th$

$Median = \frac{31}{2}th$

$Median = 15.5th$

This means that the median is the mean of the 15th and the 16th item. So,

$Median = \frac{5+5}{2}$

$Median = \frac{10}{2}$

$Median = 5$

<em>Person B has a greater median of 5</em>

Question 12: Probability that B gets 5 or 6

This is calculated as:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

From the sample space of person B, we have:

$n(5\ or\ 6) =n(5) + n(6)$

$n(5\ or\ 6) =8+10$

$n(5\ or\ 6) = 18$

So, we have:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

$P(5\ or\ 6) = \frac{18}{30}$

$P(5\ or\ 6) = 0.60$

$P(5\ or\ 6) = 60\%$

Question 13: Person with higher probability of 3 or more

Person A

$n(3\ or\ more) = 16$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}$

$P(3\ or\ more) = \frac{16}{32}$

$P(3\ or\ more) = 0.50$

$P(3\ or\ more) = 50\%$

Person B

$n(3\ or\ more) = 28$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}$

$P(3\ or\ more) = \frac{28}{30}$

$P(3\ or\ more) = 0.933$

$P(3\ or\ more) = 93.3\%$

By comparison:

$93.3\% > 50\%$

Hence, person B has a higher probability of 3 or more

7 0

3 years ago