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astraxan [27]
3 years ago
13

A football team tries to move the ball forward as many yards as possible on each play, but sometimes they end up behind where th

ey started. The distances, in yards, that a team moves on its first five plays are 2, 21, 4, 3, and 25. A positive number indicates moving the ball forward, and a negative number indicates moving the ball backward. 1). Which number in the list is the greatest? 2). What is a better question to ask to find out which play went the farthest from where the team started? 3). The coach considers any play that moves the team more than 4 yards from where they started a "big play." Which play(s) are big plays?
Mathematics
1 answer:
Illusion [34]3 years ago
3 0

Answer:

Given:

The distances, in yards, that a team moves on its first five plays are 2, 21, 4, 3, and 25.

Solved:

1. The greatest number is 25

2. If the moved distances are square, which one is largest?

3. The move which is greater than 4 is considered "big play"

=> 21 and 25 are big play (21 > 4, 25 > 4)

Hope this helps!

:)

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11) Write the explicit formula
Illusion [34]

Answer:

365

Step-by-step explanation:

tn = t1 + (n - 1)(d)

t(47) = -3 + (47 - 1)(8)

t(47) = -3 + (46)(8)

t(47) = -3 + 368

t(47) = 365

6 0
2 years ago
You spin the spinner, flip a coin, then spin the spinner again. Find the probability of the compound event. Write your answer as
Lesechka [4]

The correct question is;

You spin the spinner, flip a coin, then spin the spinner again. Find the probability of the compound event. Write your answer as a fraction or percent. If necessary round your answer to the nearest hundredth. Find the probability of spinning an odd number, flipping heads, then spinning another odd number? Write your answer as a fraction in simplest form.

Answer:

If the spinner has an even number of spaces, the probability of spinning an odd number, flipping heads, then spinning a yellow is = 1/8

Whereas, if the spinner has an odd number of spaces, more information is needed

Step-by-step explanation:

The answer will depend on whether the spinner has an odd or even number of spaces.

Now, if we suppose that the spinner has an even amount of numbers on it, there will be an equal number of odd and even numbers.

Thus the probability of spinning an odd number is ½.

Also, the probability of flipping heads on a coin is ½.

Thus, the probability of spinning an odd number, flipping heads, then spinning another odd number is;

= ½ × ½ × ½ = ⅛

Similarly,

If the spinner has an odd amount of numbers on it, the probability of spinning an odd number will vary. For instance, if there are 5 spaces on the spinner, the probability of spinning an odd number will be

3

However, if there are 9 spaces, the probability will be 5/9

So,we can conclude that;

If the spinner has an even number of spaces, the probability of spinning an odd number, flipping heads, then spinning a yellow is = 1/8

Whereas, if the spinner has an odd number of spaces, more information is needed.

6 0
2 years ago
Factor this expression completely.<br> 36y2 - 1
shtirl [24]

Answer:

(6y - 1)(6y + 1)

Step-by-step explanation:

To factorize completely, we may adopt the difference of 2 squares approach. The theory states that difference of two square  is the product of the difference of the number and the sum of the numbers.

Such that

a² - b² = (a - b) (a + b)

Hence 36y² - 1

= 6²y² - 1²

= (6y)² - 1²

= (6y - 1)(6y + 1)

5 0
3 years ago
Read 2 more answers
Thanks that helped a lot with my math homework!:):):)
frosja888 [35]
:):):):):):):):):):)
8 0
3 years ago
Each of 16 students measured the circumference of a tennis ball by four different methods, which were: A: Estimate the circumfer
almond37 [142]

Answer:

Following are the solution to the given equation:

Step-by-step explanation:

Please find the complete question in the attachment file.

In point a:

\to \mu=\frac{\sum xi}{n}

       =22.8

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{119.18}{16-1}}\\\\ =\sqrt{\frac{119.18}{15}}\\\\ = \sqrt{7.94533333}\\\\=2.8187

In point b:

\to \mu=\frac{\sum xi}{n}

       =20.6875  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{26.3375}{16-1}}\\\\=\sqrt{\frac{26.3375}{15}}\\\\ =\sqrt{1.75583333}\\\\ =1.3251

In point c:

 \to \mu=\frac{\sum xi}{n}

         =21  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{2.62}{16-1}}\\\\ =\sqrt{\frac{2.62}{15}} \\\\= \sqrt{0.174666667}\\\\=0.4179

In point d:

\to \mu=\frac{\sum xi}{n}

       =20.8375  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{8.2975}{16-1}}\\\\ =\sqrt{\frac{8.2975}{15}} \\\\  =\sqrt{0.553166667} \\\\ =0.7438

6 0
3 years ago
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