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3 years ago

Choose the best choice for the question

Mathematics

1 answer:

geniusboy [140]3 years ago

8 0

Step-by-step explanation:

a and b may be the correct answer

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Select Parameter or Statistic to classify each statement.

REY [17]

Answer:

Step-by-step explanation:

A parameter describes a population, and a statistic describes a sample.

The first one is a statistic. A sample was surveyed.

The second one is a parameter. It describes the entire soccer team.

The third one is a statistic. A sample was surveyed.

The fourth one is a parameter. It describes the entire golf team.

6 0

4 years ago

Both the La Plata river dolphin (Pontoporia blainvillei) and

Citrus2011 [14]

Answer:

1. A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin; 2. The radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball; 3. The volume of the larger ball is 3 orders of magnitude bigger than the volume of the smaller ball.

Step-by-step explanation:

If we expressed a number as:

$\\ N = a * 10^{b}$ (1)

Where

$\\ \frac{1}{\sqrt{10}} \leq a < \sqrt{10}$ (2)

$\\ 1 \leq a < 10$ (3)

Then, b represents the order of magnitude of such a number (Order of magnitude (2020), in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The La Plata river dolphin weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the scientific notation):

$\\ 30kg \leq Dolphin_{weight} \leq 50kg$

$\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg$

$\\ 35000kg \leq Whale_{weight} \leq 40000kg$

$\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg$

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

$\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)$

$\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)$

Then

$\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}$

Thus

A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin.

<h3>Second case</h3>

Following the same reasoning, we can conclude that the radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball:

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}$

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = 10^{1-0} = 10^{1}$

<h3>Third case</h3>

For this case, we need to calculate the volume of a sphere for both radii (1cm and 10cm).

The volume of a sphere is

$\\ V_{sphere} = \frac{4}{3}*\pi*R^{3}$

Then, the volume of the ball of radius 1cm is:

$\\ V_{radius=1} = \frac{4}{3}*\pi*(1cm)^{3}$

$\\ V_{radius=1} \approx 4.19*10^{0}cm^{3}$

And, the volume of the ball of radius 10cm is:

$\\ V_{radius=10} = \frac{4}{3}*\pi*(10cm)^{3}$

$\\ V_{radius=10} \approx 4.19*10^{3}cm^{3}$

Thus

$\\ \frac{10^{3}}{10^{0}} = 10^{3}$

As a result, the volume of the larger ball is 3 orders of magnitude bigger than the volume of the smaller ball.

4 0

3 years ago

Find the roots of the quadratic equation w+w²/3=0

Daniel [21]

I assume that the equation you mean is below:

$\large \boxed{w + \frac{ {w}^{2} }{3} = 0}$

To find roots for this equation, we have to get rid of the denominator. We can do by multiplying both sides by 3.

$\large{w(3) + \frac{ {w}^{2} }{3} (3) = 0(3)} \\ \large{3w + {w}^{2} = 0}$

Factor w-term out (common factor)

$\large{w(3 + w) = 0} \\ \large{w = 0 \: \: \: or \: \: \: 3 + w = 0} \\ \large{w = 0, - 3}$

Answer

The roots of quadratic equation are 0,-3

8 0

3 years ago

Find the approximate area between the curve f(x) = -4x² + 32x and on the x-axis on the interval [0,8] using 4 rectangles. Use th

Doss [256]

Split up the interval [0, 8] into 4 equally spaced subintervals:

[0, 2], [2, 4], [4, 6], [6, 8]

Take the right endpoints, which form the arithmetic sequence

$r_i=2+\dfrac{8-0}4(i-1)=2i$

where 1 ≤ i ≤ 4.

Find the values of the function at these endpoints:

$f(r_i)=-4{r_i}^2+32r_i=-16i^2+64i$

The area is given approximately by the Riemann sum,

$\displaystyle\int_0^8f(x)\,\mathrm dx\approx\sum_{i=1}^4f(r_i)\Delta x_i$

where $\Delta x_i=\frac{8-0}4=2$ ; so the area is approximately

$\displaystyle2\sum_{i=1}^4(-16i^2+64i)=-32\sum_{i=1}^4i^2+128\sum_{i=1}^4i=-32\cdot\frac{4\cdot5\cdot9}6+128\cdot\frac{4\cdot5}2=\boxed{320}$

where we use the formulas,

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$

6 0

3 years ago

If anyone could do this you be a literal life saver

mylen [45]

Answer:

Step-by-step explanation:

I really hope that solving a few trig questions can forestall death. That would be a really easy way to get rid of the C 19 restrictions and a myriad other diseases. BETTER LIFE THROUGH MATH will be our motto.

11) tan45 = x/13 x = 13

cos45 = 13/y y = 13√2

12) sin30 = 4/x x = 8

tan30 = 4/y y = 4√3

13) sin60 = 21/x x = 21(2/√3) = 42/√3

y = xcos60 = (42/√3)(1/2) = 21/√3

14) cos45 = x/22 x = 22/√2

sin45 = y/22 y = 22/√2

15) Height = 24sin30 = 12

cos 45 = 12/x x = 12/√2

tan45 = y/12 y = 12

cos30 = z/24 z = 24(√(3)/2) = 12/√3

8 0

3 years ago