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Volgvan
3 years ago
9

a soccer ball has a circumference of about 28inches while the goal is 24 feet wide how many soccer balls would be needed to cove

r the distance between the goal post
Mathematics
2 answers:
Zielflug [23.3K]3 years ago
7 0
We know that
[the circumference of a circle]=pi*D
D=circumference/pi-----------------> 28/pi-----------------> D=8.91 in

1 ft---------> 12 in
X ft--------> 8.91 in
X=8.91/12------> X=0.7425 ft

then
24/0.7425--------> 32.32 soccer balls

the answer is
33 balls soccer

scoray [572]3 years ago
4 0
To find the number of soccer balls that would be needed to cover the distance to the goal, you will need to find the diameter of the soccer ball.

To find the diameter use be formula for circumference to solve for the diameter.

C = pi x d
28 = 3.14 x d
d = 8.9 inches

The distance to cover is 24 feet, so convert to inches and divide by 8.9 inches.

24 x 12 = 288 inches
288/8.9 = 32.4 balls
You will need about 32 soccer balls to cover the distance.
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If a quadratic equation with real coefficients has a discriminant of 3 then the two roots must be
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3 years ago
Whats the slope of (8,18) and (16,24)
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3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%20%2B%201%7D%7B2x%20-%201%7D%20%20-%20%20%5Cfrac%7Bx%20-%201%7D%7B2x%20%2B%201%
lubasha [3.4K]

x=0

because you must

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8 0
2 years ago
The following data lists the ages of a random selection of actresses when they won an award in the category of Best​ Actress, al
Valentin [98]

Answer:

a) p_v =P(t_{(9)}

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best​ Actors at 1% of significance.

b) The 99% confidence interval would be given by (-21.469;2.069)

c) We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two

Step-by-step explanation:

Part a

Let put some notation  

x=actor's age , y = actress's age

x: 58 41 36 36 34 33 48 37 37 43

y: 26 27 34 26 35 29 23 42 30 34

The system of hypothesis for this case are:

Null hypothesis: \mu_y- \mu_x \geq 0

Alternative hypothesis: \mu_y -\mu_x

The first step is calculate the difference d_i=y_i-x_i and we obtain this:

d: -32, -14, -2, -10, 1, -4, -25, 5, -7, -9

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= -9.7

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =11.451

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-9.7 -0}{\frac{11.451}{\sqrt{10}}}=-2.679

The next step is calculate the degrees of freedom given by:

df=n-1=10-1=9

Now we can calculate the p value, since we have a left tailed test the p value is given by:

p_v =P(t_{(9)}

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best​ Actors at 1% of significance.

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The confidence interval for the mean is given by the following formula:  

\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that t_{\alpha/2}=3.25  

Now we have everything in order to replace into formula (1):  

-9.7-3.25\frac{11.451}{\sqrt{10}}=-21.469  

-9.7+3.25\frac{11.451}{\sqrt{10}}=2.069  

So on this case the 99% confidence interval would be given by (-21.469;2.069)

Part c

We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two means is 0.

8 0
3 years ago
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