All categories

3 years ago

A) find a cartesian equation for the curve described by the given polar equation. r =3 sin g

1 answer:

6 0

Recall that r^2 = x^2 + y^2, so that r = sqrt(x^2+y^2).

y
r = 3 sin g becomes sqrt(x^2+y^2) = 3*-----------------------
sqrt(x^2+y^2)

Squaring both sides,

9y^2
x^2+y^2 = -----------------
x^2 + y^2

If this is correct (and I'm not convinced that it is), then (x^2+y^2)^2 = 9y^2
shows the relationship between x and y. Can anyone improve on this result?

You might be interested in

Pls help 20 points and mark brainliest

tatiyna

1:equilateral
2:isosceles
3:scalene
4:right
5:acute
6:obtuse

8 0

3 years ago

Read 2 more answers

The burges are moving across the country .Mr burges leaves 3 hours before mrs.burges. If he averages 50 mph and she averages 70

kirill [66]

10.5 hours

let t be time then distance travelled by both is

Mr Burges ⇔ d = 50t ( distance = speed × time )

Mrs Burges ⇒ d = 70(t - 3)

At the point of overtaking both will have travelled the same distance

equating their distance and solving for t

70(t - 3) = 50t

70t - 210 = 50t ( subtract 50t from both sides )

20t - 210 = 0 ( add 210 to both sides )

20t = 210 ( divide both sides by 20 )

t = 10.5 hours

4 0

3 years ago

Suppose the mean SAT verbal score is 525 with a standard deviation of 100, while the mean SAT math score is 575 with a standard

Ipatiy [6.2K]

Answer:

The mean of the combined math and verbal scores is 1100, while the standard deviation is 141.

Step-by-step explanation:

Normal variables

Normal variables have mean $\mu$ and standard deviation $\sigma$

When we add normal variables, the combined mean is the sum of both means, and the standard deviation is the square root of the sum of both variances. The distribution is still normal.

In this question:

Verbal: $\mu_{V} = 525, \sigma_{V} = 100$ .

Math: $\mu_{M} = 575, \sigma_{M} = 100$

Combined:

$\mu = \mu_{V} + \mu_{M} = 525 + 575 = 1100$

$\sigma = \sqrt{\sigma_{V}^{2}+\sigma_{M}^{2}} = \sqrt{100^2 + 100^2} = 141$

The mean of the combined math and verbal scores is 1100, while the standard deviation is 141.

6 0

3 years ago

33. sayfa yaparmısınız

zalisa [80]

Answer:

hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

8 0

3 years ago

One sample has a mean of and a second sample has a mean of . The two samples are combined into a single set of scores. What is t

Murrr4er [49]

Answer:

a) For this case we can use the definition of weighted average given by:

$M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}$

And if we replace the values given we have:

$M = \frac{8*4 + 16*4}{4+4}= 12$

b) $M = \frac{8*3 + 16*5}{3+5}= 13$

c) $M = \frac{8*5 + 16*3}{5+3}= 11$

Step-by-step explanation:

Assuming the following question: "One sample has a mean of M=8 and a second sample has a mean of M=16 . The two samples are combined into a single set of scores.

a) What is the mean for the combined set if both of the original samples have n=4 scores "

For this case we can use the definition of weighted average given by:

$M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}$

And if we replace the values given we have:

$M = \frac{8*4 + 16*4}{4+4}= 12$

b) what is the mean for the combined set if the first sample has n=3 and the second sample has n=5

Using the definition we have:

$M = \frac{8*3 + 16*5}{3+5}= 13$

c) what is the mean for the combined set if the first sample has n=5 and the second sample has n=3

Using the definition we have:

$M = \frac{8*5 + 16*3}{5+3}= 11$

7 0

3 years ago