I am confused. what is 5r

An amusement park offers an entrance fee discount of 20% to children under 6 years of age and seniors 65 or older. What is the r

Temka [501]

Answer:

[0, 6) ∪ [65, ∞)

Step-by-step explanation:

Ages under 6 will be in the interval [0, 5], or in the interval [0, 6), as you may wish.

Ages 65 or older will be in the interval [65, ∞).

The union of these two intervals is ...

[0, 6) ∪ [65, ∞)

5 0

3 years ago

Denise spent $3.45 on snacks every day for 11 days. What is the amount of money Denise spent on these snacks?

Ipatiy [6.2K]

Answer:

Your Answer would be C: $37.95

Step-by-step explanation:

Take 3.45 Multiply by 11 And get your answer

6 0

3 years ago

Read 2 more answers

37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the

Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy$

and I assume C has a positive orientation in both cases

(a) It looks like the region has the curves y = x and y = x ² as its boundary***, so that the interior of C is the set D given by

$D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}$

• Compute the line integral directly by splitting up C into two component curves,

C₁ : x = t and y = t ² with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}$

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}$

(b) C is the boundary of the region

$D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}$

• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = t with 0 ≤ t ≤ 1

C₃ : x = 0 and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}$

• Using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}$

4 0

3 years ago

PLEASE HELP!! WILL MARK BRAINLIEST!!!

Rus_ich [418]

Answer:

that looks tuff my boy good luck

Step-by-step explanation:

7 0

3 years ago

The mean percentange of a population of people eating out at least once a week is 57℅

Sidana [21]

Answer:

The correct answer is B. between 56.45% and 57.55%

Complete statement and question:

The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.50%. Assume that a sample size of 40 people was surveyed from the population a significant number of times. In which interval will 68% of the sample means occur?

between 55.89% and 58.11%

between 56.45% and 57.55%

between 56.54% and 57.46%

between 56.07% and 57.93%

Source: brainly.com/question/1068489

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Mean percentage of a population of people eating out at least once a week = 57%

Standard deviation = 3.5%

Sample size = 40

Confidence level = 68%

2. In which interval will 68% of the sample means occur?

For answering this question, we should find out the standard deviation of the sample, using this formula:

Standard deviation of the sample = Standard deviation of the population/√Sample size

Standard deviation of the sample = 3.5/√40

Standard deviation of the sample = 3.5/6.32

Standard deviation of the sample = 0.55

Let's recall that a confidence level of 68% means that 68% of the sample data would have a value between the mean - 1 time the standard deviation of the sample and the mean + 1 time the standard deviation of the sample. Thus:

57 - 1 * 0.55 = 57 - 0.55 = 56.45

57 + 1 * 0.55 = 57 + 0.55 = 57.55

The correct answer is B. between 56.45% and 57.55%

7 0

3 years ago

I am confused. what is 5r– 7 = 2r+ 14