Anyone know how to do cot^6x=cot^4csc^2x-cot^4x

The sun is about 93 * 10^6 miles from earth.What is this distance written as a whole number?

Sergio039 [100]

93,000,000

$93 \times 10 ^{6} = 93000000$

8 0

3 years ago

Tim Worker expects Social Security benefits of $937 per month. His wife, Mary, is eligible for a wife's benefit of 50 percent of

mihalych1998 [28]

Answer:

468.5

Step-by-step explanation:

so that would be $468 and 50 cents

8 0

3 years ago

In a random sample of 80 teenagers, the average number of texts handled in a day is 50. The 96% confidence interval for the mean

Nastasia [14]

Answer:

a) $\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

b) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

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".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n=80 represent the sample size

Solution to the problem

Part a

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

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

For this case we can calculate the mean like this:

$\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

Part b

For this case is the sample size is doubled the margin of error would be:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

5 0

3 years ago

The total number of restaurant-purchased meals that the average person will eat in a restaurant, in a car, or at home in a ye

zloy xaker [14]

Answer:

Number of restaurant-purchased meals eaten in a restaurant = 70

Number of restaurant-purchased meals eaten in a car = 22

Number of restaurant-purchased meals eaten in a home = 60

Step-by-step explanation:

Let the number of people that will eat in a restaurant be R.

Let the number of people that will eat in a car be C.

Let the number of people that will eat in a home be H.

From the information given in the problem we have,

R+C+H = 152 ____equation (1)

C+H-R = 12 ____equation (2)

R = H+10 ____equation (3)

1) Plugging in R=H+10 from equation 3 into the equation 2, we get

C+H-R=12

=> C+H-(H+10)=12

=> C+H-H-10=12

2) Cancelling out +H and -H, we get

C-10=12

3) Add 10 to both sides

C-10+10=12+10

4) Cancelling out -10 and +10, we get

C=22

5) Plugin C=22 in equation 1, we have

R+C+H = 152

=> R+22+H=152

Subtract 22 from both sides,

R+22+H-22=152-22

Cancelling out +22 and -22 from the left side, we get

R+H=130 ____let it be equation (4)

6) Plugin C=22 in equation 2, we have

C+H-R = 12

22+H-R = 12

Subtracting 22 from both the sides, we get

22+H-R-22 = 12-22

Cancelling out +22 and -22 from the left side,

H-R = -10 ____let it be equation (5)

7) Adding equation 4 and equation 5, we get

(R+H)+(H-R) = 130 + (-10)

=> R+H+H-R = 130-10

8) Cancelling out R and -R from the left side, we get

2H = 120

9) Dividing both sides by 2, we get

$\frac{2H}{2} = \frac{120}{2}$

10) Cancelling out the 2's from the left side, we have

H=60

11) Plugging in C=22 and H=60 in the equation 1, we have

R+C+H = 152

=> R+22+60=152

=> R + 82 = 152

12) Subtracting 82 from both the sides, we get

R + 82 -82 = 152 -82

13) Cancelling out +82 and -82 from the left side, we get

R = 70

8 0

3 years ago

How much higher is 500 than -400 meters

aleksandr82 [10.1K]

I believe 900 meters your welcomes

7 0

3 years ago

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