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3 years ago

From end zone to end zone, a football field is 120 yards long. How long is the football field in feet?

Mathematics

2 answers:

amm18123 years ago

7 0

Answer:

It's simple the answer is 360

Step-by-step explanation:

cricket20 [7]3 years ago

4 0

Answer:

The football field is 360 feet long.

Step-by-step explanation:

Remember, there are 3 feet in 1 yard.

1 yard = 3 feet

To find out how long the football field in feet, we must convert 120 yards into feet.

To convert yards to feed, multiply the number of yards by 3 (3 feet per yard).

120 × 3 = 360

The football field is 360 feet long.

I hope this helps. :)

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SAT scores: The College Board reports that in 2009, the mean score on the math SAT was 582 and the population standard deviation

Andrews [41]

Answer:

we can conclude that there is no significant evidence to conclude that the mean score in 2010 differs from the mean score in 2009.

Step-by-step explanation:

H0 : μ = 582

H1 : μ < 582

Test statistic :

T = (xbar - μ) ÷ σ/√n

Xbar = 515 ; n = 20 ; σ = 120

T = (515 - 582) ÷ 120/√20

T = -67 / 26.832815

T = 2.50

Pvalue at t score = 2.50 ; df = 19 is 0.0187

At α = 0.0187

Pvalue > α ; Hence, we fail to reject the Null

Hence, we can conclude that there is no significant evidence to conclude that the mean score in 2010 differs from the mean score in 2009.

7 0

3 years ago

Amelia receives two dozens roses for her birthday. 18 of the roses are pink what percent of the roses are pink

ollegr [7]

18/24 = pink roses... 75% of the roses are pink.

4 0

3 years ago

he number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.02

Yakvenalex [24]

Answer:

a) 98.01%

b) 13.53\%

c) 27.06%

Step-by-step explanation:

Since a car has 10 square feet of plastic panel, the expected value (mean) for a car to have one flaw is 10*0.02 = 0.2

If we call P(k) the probability that a car has k flaws then, as P follows a Poisson distribution with mean 0.2,

$P(k)= \frac{0.2^ke^{-0.2}}{k!}$

In this case, we are looking for P(0)

$P(0)= \frac{0.2^0e^{-0.2}}{0!}=e^{-0.2}=0.9801=98.01\%$

So, the probability that a car has no flaws is 98.01%

Ten cars have 100 square feet of plastic panel, so now the mean is 100*0.02 = 2 flaws every ten cars.

Now P(k) is the probability that 10 cars have k flaws and

$P(k)= \frac{2^ke^{-2}}{k!}$

and

$P(0)= \frac{2^0e^{-2}}{0!}=0.1353=13.53\%$

And the probability that 10 cars have no flaws is 13.53%

Here, we are looking for P(1) with P defined as in b)

$P(1)= \frac{2^1e^{-2}}{1!}=2e^{-2}=0.2706=27.06\%$

Hence, the probability that at most one car has no flaws is 27.06%

6 0

3 years ago

Read 2 more answers

55% of what number is 231

babunello [35]

.55*X=231 .....X=231/.55.......X= 420

4 0

3 years ago

Which is the approximate solution to the system y = 0.5x + 3.5 and y = − 2/3 x + 1/3 shown on the graph? (–2.7, 2.1) (–2.1, 2.7)

AlekseyPX

Answer:

The approximate solution to the system is $(-2.7, 2.1)$ .

Step-by-step explanation:

To solve the system of equations $\begin{bmatrix}y=0.5x+3.5\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$ you must:

$\mathrm{Rationalize\:equations}\\\\\begin{bmatrix}y=\left(\frac{1}{2}\right)x+\left(\frac{7}{2}\right)\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$

$\mathrm{Subsititute\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\begin{bmatrix}-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\\\\-\frac{2}{3}x=\frac{19}{6}+\frac{1}{2}x\\\\-\frac{7}{6}x=\frac{19}{6}\\\\6\left(-\frac{7}{6}x\right)=\frac{19\cdot \:6}{6}\\\\-7x=19\\\\x=-\frac{19}{7}\approx-2.7$

$\mathrm{For\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\mathrm{Subsititute\:}x=-\frac{19}{7}\\\\y=-\frac{2}{3}\left(-\frac{19}{7}\right)+\frac{1}{3}\\\\y=\frac{15}{7}\approx 2.1$

The approximate solutions to the system of equations are:

$x=-2.7 ,\:y=2.1$

5 0

3 years ago

Read 2 more answers