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AVprozaik [17]
2 years ago
12

Your band is one of 8 bands competing at the battle of the bands. The order of the performances is determined at random. The fir

st 5 performances are on Friday night and the next 3 are on Saturday.
What is the probability that your band gives the last performance on Friday night and your rival band performs immediately before you?
Mathematics
1 answer:
lorasvet [3.4K]2 years ago
3 0

Answer:

1.79 %.

Step-by-step explanation:

you have to calculate the probability of a series of events before you can calculate the final probability.

The first event is that the band plays on Friday.

If there are 8 bands, but only 5 play on Friday, then the probability that they will play on Friday is

5/8

Now, the second event that they are the last ones is 1/5, since they are 5 bands.

Therefore the probability that they will play on Saturday and last is

5/8 * 1/5 = 1/8

Now, for the rival band it would be, knowing that on Friday there is already a quota assigned and one band less, the probability that they will play that day is

4/7

as they play before the other group, it would be 1/4, as there is already a quota assigned.

The final probability would then be:

4/7 * 1/4 = 1/7

Now the probability that the band gives the last performance on Friday night and your rival band performs immediately before them, is:

1/8 * 1/7 = 0.0179

That is 1.79 %.

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Camille knew that a triangle had one side with a length of 16 inches and another side with a length of 20 inches. She did not kn
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The unknown side is 9. Because 9 x 5= 45, 45 being the perimeter, that is 5 times the unknown side. And 45 is the sum of 9, 16, and 20.
5 0
2 years ago
It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the
JulijaS [17]

Answer:

1)  n=48  

2) n=298

3) n=426

Step-by-step explanation:

Previous concepts

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".  

p represent the real population proportion of interest

\hat p represent the estimated proportion for the sample

n is the sample size required (variable of interest)

z represent the critical value for the margin of error

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.90=0.10 and \alpha/2 =0.05. And the critical value would be given by:  

z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64  

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}} (a)  

And on this case we have that ME =\pm 0.1 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:  

n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63  

And rounded up we have that n=48  

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7  

And rounded up we have that n=298  

Part 3

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:  

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96  

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}} (a)  

And on this case we have that ME =\pm 0.04 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:  

n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.96})^2}=425.22  

And rounded up we have that n=426  

3 0
2 years ago
Your utility company charges 11 cents per​ kilowatt-hour of electricity. Complete parts​ (a) and​ (b) below. a. What is the dail
Colt1911 [192]

Answer:

a. 9.9 cents

b. 26.5 dollars

Step-by-step explanation:

The first thing they mention to us is the cost of energy 11 cents / kwh

a) In the first point they ask us to calculate the daily cost of keeping the flashlight on for 12 hours, they tell us that it consumes 75 watts per hour, but this value must be passed to kW because the cost is given in kWh

to pass it, it would be: 75 watt * 1 kw / 1000 watt = 0.075 kw,

the formula to apply would be like this:

Daily cost = flashlight cost * time * energy cost

we all know them so we replace:

Daily cost = 0.075 kw * 12 h * 11 cents / kwh = 9.9 cents

b) Now, to know the annual savings, we must calculate the difference between the expense of the normal flashlight and the LED, let's calculate the daily cost of the LED, like this:

20 watt * 1 kw / 1000 watt = 0.02 kw

Daily cost = 0.02 kw * 12 h * 11 cents / kwh = 2.64 cents

Now by calculating the difference we multiply by 365 which are the days of a year.

Savings / year = (9.9 - 2.64) * 365 = 2649.9 cents = 26.5 dollars.

5 0
2 years ago
Which shows the correct first step to solving the system of equations in the most efficient manner? 3 x + 2 y = 17. x + 4 y = 19
skelet666 [1.2K]

Answer: First option.

Step-by-step explanation:

<h3> The missing picture is attached.</h3><h3></h3>

You can use these methods to solve a System of Equations and find the value of the variables:

A. Substitution method.

B. Elimination method.

In this case, given the following System of equations:

\left \{ {{3 x + 2 y = 17} \atop {x + 4 y = 19}} \right.

The most efficient method to solve it is the Substitution Method. The procedure is:

Step 1: Solve for one of the variables from the most convenient equation.

Step 2: Make the substitution into the other equation and solve for the other variable in order to find its value.

Step 3: Substitute the value obtained into the equation from Step 1 and evaluate, in order to find the value of the  other variable.

Based on this, you can identify that first step to solve the given System of equations, is solving for "x" from the second equation:

x + 4 y = 19\\\\x=-4y+19

3 0
2 years ago
Read 2 more answers
se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y
alexandr1967 [171]

Answer:

It is proved that f_x, f_y exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)

  • To show exixtance of f_x(0,0), f_y(0,0) we take,

f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0    exists.

And,

f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}   exists.

  • To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}  where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

4 0
2 years ago
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