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3 years ago

Which is the best estimate for (8.9x10^8)/(3.3x10^4) written in scientific notation? A. 3x10^2 B. 6x10^2 C. 3x10^4 D. 6x10^4

Mathematics

2 answers:

xenn [34]3 years ago

7 0

Esitmate:
9 /3 = 3
and
8-4 = 4
so it would be 3x10^4
answer is C

son4ous [18]3 years ago

3 0

Answer:

Option C

Step-by-step explanation:

The given expression is $\frac{8.9\times 10^{8}}{3.3\times 10^{4}}$

Now we have to solve this.

$\frac{8.9\times 10^{8}}{3.3\times 10^{4}}$

$=\frac{8.9}{3.3}\times \frac{10^{8}}{10^{4}}$

= 2.70 × 10⁽⁸⁻⁴⁾

= 2.70 × 10⁴

≈ 3 × 10⁴

Therefore, Option C is the answer.

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3 years ago

I need help on my homework

soldi70 [24.7K]

Answer:

$\displaystyle m\angle AED=32.5^\circ$

Step-by-step explanation:

Angles in a Circle

An exterior angle of a circle is an angle whose vertex is outside a circle and the sides of the angle are secants or tangents of the circle.

Segments AE and DE are secants of the given circle. They form an exterior angle called AED.

The measure of an exterior angle is equal to half the difference of the measure of their intercepted arcs.

Intercepted arcs in the given circle are AD=113° and BC=48°. The exterior angle is:

$\displaystyle m\angle AED=\frac{AD-BC}{2}$

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8 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

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

• 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

What is the confidence interval estimate of the population mean percentage change in the price per share of stock during the fir

I am Lyosha [343]

Answer:

(-0.1059 ; - 0.0337)

Step-by-step explanation:

The data table is attached in the picture below:

These is a matched pair design ; which requires taking the difference of the two values for each sample :

The mean and standard deviation of the difference will be used to construct the confidence interval :

The mean of difference, dbar = Σx/n = - 0.0698

The standard deviation of difference, Sd ;

Sd = [√Σ(d - dbar)²/(n-1)] = 0.1054

n = sample size = 25

The confidence interval :

dbar ± [TCritical * Sd/√n]

Tcritical at 90% ; df = n -1 = 25 -1

Tcritical(90% , 24) = 1.1711

C.I = - 0.0698 ± (1.711 * 0.1054/√25)

C.I = - 0.0698 ± 0.0361

C.I = (-0.1059 ; - 0.0337)

5 0

3 years ago

You have a garden in your backyard that measures 15 meters by 20 meters. You want to place stepping stones around the perimeter

Anettt [7]

Answer:

0.5 m

Step-by-step explanation:

Given that the dimensions of the garden are 15 meters by 20 meters.

The area of the new garden (original garden + stone placed around the perimeter) is 336m².

Let d be the width of the border having stepping stones as shown in the figure. The shaded region in the figure is the area having stepping stones.

The area including the shaded region $= (15+2d)(20+2d) m^2$

$\Rightarrow 336 = (15+2d)(20+2d) \\\\\Rightarrow 336 = 300+ 70 d +4d^2 \\\\\Rightarrow 4d^2 +70d+300-336=0 \\\\\Rightarrow 4d^2 +70d-36=0 \\\\\Rightarrow 2d^2 +35d-18=0 \\\\\Rightarrow 2d^2 +36d-d-18=0 \\\\\Rightarrow (2d-1)(d+18)=0 \\\\\Rightarrow (2d-1)=0, \; or \; (d+18) \\\\\Rightarrow d= 0.5\;or -18 \\\\$

As the width of the stone border can't be a negative value, so taking the positive value.

Hence, the width of the stone border is 0.5 m.

So, the wi

7 0

3 years ago