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3 years ago

PWEASE HELP ASAP my friends and I cant figure this out I'm hoping one of u guys can

Mathematics

1 answer:

Degger [83]3 years ago

3 0

The number of viruses was 4000 times greater than number of protozoa.

Step-by-step explanation:

Given,

Number of protozoa = $5*10^5$

Number of viruses = $2*10^9$

Let,

x be the number of times number of viruses is greater than number of protozoa.

Number of viruses = Number of time * Number of protozoa

$2*10^9=(5*10^5)x$

Dividing both sides by $5*10^5$

$\frac{2*10^9}{5*10^5}=\frac{(5*10^5)x}{5*10^5}\\\\4*10^3= x\\x=4*10^3 \\x=4000$

The number of viruses was 4000 times greater than number of protozoa.

Keywords: division, scientific notation

Learn more about division at:

#LearnwithBrainly

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MATH: EXPONENTIAL WORK PROBLEM 1. HELP PLEASE! 13 pts

Strike441 [17]

The amount of the radioactive substance is 374.6 g

<h3>How to determine the amount of radioactive substance?</h3>

The given parameters are:

Initial, a = 424 mg
Rate, r = 6%
Time, t = 2 hours

The amount of the radioactive substance is calculated as:

A(t) = a(1 - r)^t

This gives

A(t) = 424 * (1 - 6%)^t

At 2 hours, we have:

A(2) = 424 * (1 - 6%)^2

Evaluate

A(2) = 374.6

Hence, the amount of the radioactive substance is 374.6 g

Read more about exponential functions at:

brainly.com/question/2456547

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4 0

2 years ago

If two standard six-sided dice are tossed, what is the probability that a 5 is rolled on at least one of the two dice? express y

zaharov [31]

We define the probability of a particular event occurring as:
$\frac{number\ of \ desired\ outcomes}{number\ of\ possible\ outcomes}$

What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are <em>independent</em>, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for <em>each </em>of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.

Now that we've found the number of possible outcomes, we need to find the number of <em>desired</em> outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is <em>at least one 5 rolled</em>. It turns out, there are only 3:

(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5

So, we have $\frac{3}{36} = \frac{1}{12}$ probability of rolling at least one 5.

6 0

3 years ago

use coordinate notation to write the rule that maps each preimage to its image. The identify the transformation and confirm that

Ymorist [56]

The <em>rigid</em> transformations used for each figure:

Figure 5 - Reflection around x and y axes: (x, y) → (- x, - y)
Figure 6 - Horizontal and vertical translations: (x, y) → (x + 1, y - 2)

<h3>What transformation rules do create the resulting images?</h3>

In this question we must determine what kind of <em>rigid</em> transformations generates each image. <em>Rigid</em> transformations are transformations applied on geometric loci such that <em>Euclidean</em> distance is conserved. Now we proceed to determine the transformation rule for each case:

Figure 5 - Reflection around the x-axis followed by reflection around the y-axis.

(x, y) → (- x, - y)

Figure 6 - Translation one unit in the +x direction and two units in the -y direction.

(x, y) → (x + 1, y - 2)

To learn more on transformation rules: brainly.com/question/9201867

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8 0

2 years ago

The figure is made up of a cylinder and a sphere which has been cut in half. The radius of each half sphere is 5 mm. What is

alexgriva [62]

<h2>Answer:</h2>

<em>Rounded to the nearest hundredth the volume of the composite figure is:</em>

<em>1308 33 cubic millimeters</em>

<h2>Explanation:</h2>

Hello! I wrote the complete question in a comment above. The volume of a cylinder is defined as:

$V_{c}=\pi r^2 h \\ \\ r:radius \\ \\ h:height$

While the volume of half a sphere is:

$V_{hs}=\frac{2}{3}\pi r^3$

Since we have 2 half spheres, then the volume of these is the same as the volume of a sphere:

$V_{s}=\frac{4}{3}\pi r^3$

Then, the composite figure is:

$V=\pi r^2 h +\frac{4}{3}\pi r^3 \\ \\ V=\pi r^2(h+\frac{4}{3}r)$

The radius of the cylinder is the same of the radius of each half sphere. So:

$r=5mm \\ \\ h=10mm \\ \\ \\ V=(3.14) (5)^2((10)+\frac{4}{3}(5)) \\ \\ V=25(3.14)(10+\frac{20}{3}) \\ \\ \boxed{V\approx 1308.33mm^3}$

7 0

3 years ago

Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat

yulyashka [42]

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

$P(t) = 0.8t^{2} + 6t + 19000$

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

$P(t) = 0.8t^{2} + 6t + 19000$

$0.8t^{2} + 6t + 19000 = 34200$

$0.8t^{2} + 6t - 15200 = 0$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$0.8t^{2} + 6t - 15200 = 0$

So $a = 0.8, b = 6, c = -15200$

Then

$\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356$

$t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14$

$t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64$

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

6 0

2 years ago