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Studentka2010 [4]
3 years ago
14

How does probability apply to basketball?

Mathematics
1 answer:
OleMash [197]3 years ago
8 0
Depending on the player, a spectator can figure out the chance of the player making the basket :D
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Find the volume of the cone. Round your answer to the nearest hundredth.
noname [10]

Answer:

can we get the figure plz

7 0
3 years ago
Convert 0.815 into a fraction in the simplest form. Explain please.
anastassius [24]

Answer:

815/1000=163/200 is the simplest form.

8 0
3 years ago
What’s the equation for this table? Use y=mx+b. HELP
Verdich [7]

Answer:

y = 2x + 12

Step-by-step explanation:

You can check this by plugging any of the values from the table into the equation. You can find the answer by finding the slope and then using that to find the y-intercept.

y2-y1/x2-x1  

16-14/2-1 = 2/1 = 2. So our slope equals 2.

If we set x = 0 to find our y-intercept, and we know that we have a slope of 2 we can subtract 2 from 14 to get the y-intercept.

14-2 = 12. --> (0, 12) which is true because we know that the slope is 2.

3 0
3 years ago
A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.
nika2105 [10]

Answer:

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

\bar X=2.55 represent the sample mean for the late time for a flight

\mu population mean

\sigma=12 represent the population deviation

n=76 represent the sample size  

Confidence interval

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

The confidence interval for the true mean is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The Confidence level given is 0.95 or 95%, th significance would be \alpha=0.05 and \alpha/2 =0.025. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got z_{\alpha/2}=1.96

Replacing we got:

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0
3 years ago
On a coordinate plane, kite W X Y Z is shown. Point W is at (negative 3, 3), point X is at (2, 3), point Y is at (4, negative 4)
xz_007 [3.2K]

Answer:

P = 10 + 2\sqrt{53} units

Step-by-step explanation:

Given

Shape: Kite WXYZ

W (-3, 3),  X (2, 3),

Y (4, -4),  Z (-3, -2)

Required

Determine perimeter of the kite

First, we need to determine lengths of sides WX, XY, YZ and ZW using distance formula;

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

For WX:

(x_1, y_1)\ (x_2,y_2) = (-3, 3),\ (2, 3)

WX = \sqrt{(-3 - 2)^2 + (3 - 3)^2}

WX = \sqrt{(-5)^2 + (0)^2}

WX = \sqrt{25}

WX = 5

For XY:

(x_1, y_1)\ (x_2,y_2) = (2, 3)\ (4,-4)

XY = \sqrt{(2 - 4)^2 + (3 - (-4))^2}

XY = \sqrt{-2^2 + (3 +4)^2}

XY = \sqrt{-2^2 + 7^2}

XY = \sqrt{4 + 49}

XY = \sqrt{53}

For YZ:

(x_1, y_1)\ (x_2,y_2) = (4,-4)\ (-3, -2)

YZ = \sqrt{(4 - (-3))^2 + (-4 - (-2))^2}

YZ = \sqrt{(4 +3)^2 + (-4 +2)^2}

YZ = \sqrt{7^2 + (-2)^2}

YZ = \sqrt{49 + 4}

YZ = \sqrt{53}

For ZW:

(x_1, y_1)\ (x_2,y_2) = (-3, -2)\ (-3, 3)

ZW = \sqrt{(-3 - (-3))^2 + (-2 - 3)^2}

ZW = \sqrt{(-3 +3)^2 + (-2 - 3)^2}

ZW = \sqrt{0^2 + (-5)^2}

ZW = \sqrt{0 + 25}

ZW = \sqrt{25}

ZW = 5

The Perimeter (P) is as follows:

P = WX + XY + YZ + ZW

P = 5 + \sqrt{53} + \sqrt{53} + 5

P = 5 + 5 + \sqrt{53} + \sqrt{53}

P = 10 + 2\sqrt{53} units

6 0
3 years ago
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