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nirvana33 [79]
3 years ago
9

Prove algebraically that r = 10/2+2sinTheta is a parabola

Mathematics
1 answer:
Xelga [282]3 years ago
4 0

Answer:

y =  -  \frac{ 1 }{10} {x}^{2}   +  \frac{5}{2}

Step-by-step explanation:

We want to prove algebraically that:

r =  \frac{10}{2 + 2 \sin \theta}

is a parabola.

We use the relations

{r}^{2}  =  {x}^{2}  +  {y}^{2}

and

y = r \sin \theta

Before we substitute, let us rewrite the equation to get:

r(2 + 2 \sin \theta) = 10

Or

r(1+  \sin \theta) = 5

Expand :

r+  r\sin \theta= 5

We now substitute to get:

\sqrt{ {x}^{2}  +  {y}^{2} }  + y = 5

This means that:

\sqrt{ {x}^{2}  +  {y}^{2} }=5 - y

Square:

{x}^{2}  +  {y}^{2} =(5 - y)^{2}

Expand:

{x}^{2}  +  {y}^{2} =25 - 10y +  {y}^{2}

{x}^{2}  =25 - 10y

{x}^{2}  - 25 =  - 10y

y =  -  \frac{ {x}^{2} }{10}  +  \frac{5}{2}

This is a parabola (0,2.5) and turns upside down.

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Tim runs 3 miles every 45 minutes. At this rate, how long did it take Tim to run 18 miles?
geniusboy [140]

Answer: 270 minutes or 4 hours and 30 minutes

Step-by-step explanation: 3x6 =18  and 45 x 6= 270

7 0
3 years ago
Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72
Lisa [10]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that \mu = 38.72, \sigma = 3.17

Sample of 10:

This means that n = 10, s = \frac{3.17}{\sqrt{10}}

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}

Z = 1.28

Z = 1.28 has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

\mu = 266, \sigma = 16

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{260 -  266}{16}

Z = -0.375

Z = -0.375 has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now n = 20, so:

Z = \frac{X - \mu}{s}

Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}

Z = -1.68

Z = -1.68 has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now n = 50, so:

Z = \frac{X - \mu}{s}

Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}

Z = -2.65

Z = -2.65 has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that n = 15. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

Z = \frac{X - \mu}{s}

Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}

Z = 2.42

Z = 2.42 has a p-value of 0.9922.

X = 256

Z = \frac{X - \mu}{s}

Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}

Z = -2.42

Z = -2.42 has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

8 0
3 years ago
Exercise No. 1
Leya [2.2K]

Answer:

don't get it I don't know how to do it

8 0
3 years ago
Please help i got a f on it and i am redoing it i will give 28 points for the right answer
Savatey [412]

Answer:

11/16

Step-by-step explanation:

12 divided by (2+2/3)^2 =1.6875

1.6875 in a decimal form is 11/16

5 0
3 years ago
Read 2 more answers
the ratio of lions supporters to sharks supporters in a sports bar is 2:5. If there are 35 people in the bar how many are sharks
Rudik [331]

Answer:

10

Step-by-step explanation:

2+5 = 7

35/7 = 5

2 × 5 = 10

hope this helps...

3 0
3 years ago
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