Simplify 23.85 divided by 0.15

What effect does decreasing the temperature of a room have on a lifespan of a battery?

Leni [432]

This question is incomplete. There are options in the question (from reference) and no highlight in the red bolded text in the question.

What effect does decreasing the temperature of a room have on a lifespan of a battery?

Which variable is highlighted in red, bolded tex?

a . Independent

b . Dependent

c . Control

d . None

But the answer to both the independent and dependent variables will be provided and explained in detail.

Answer:

Independent variable: decrease in temperature

Dependent variable: lifespan of battery

Step-by-step explanation:

The independent variable in an experiment is defined as the variable that can be changed or manipulated in order to find an outcome while the dependent variable is defined as the variable that is influenced by the changes in the independent variable.

In the given example, a decrease in temperature is the dependent variable as it is changing and the lifespan of battery is the dependent variable as it is responding to the independent variable ?9decrease in temperature).

8 0

2 years ago

2.38 Baggage fees: An airline charges the following baggage fees: $25 for the first bag and $30 for the second. Suppose 54% of p

Anna71 [15]

The average baggage-related revenue per passenger is $16.30 per passenger.

<h3>Expected value</h3>

Expected value formula: x×p(x)

First step

No passenger=0×.54

No passenger=0

Second step

One checked luggage for first bag=.30×$25

One checked luggage for first bag=$7.50

Third step

Two piece for the first and second bag=.16×($25+$30)

Two piece for the first and second bag=.16×$55

Two piece for the first and second bag=$8.80

Last step

Expected value=$7.50+$8.80

Expected value=$16.30

Therefore the average baggage-related revenue per passenger is $16.30 per passenger.

Learn more about expected value here:brainly.com/question/24305645

#SPJ1

7 0

2 years ago

Which statement about g(x)=x^2-576 is true

goldenfox [79]

We can answer without the picture or answer options

5 0

3 years ago

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

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