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Hoochie [10]
1 year ago
10

The decay of 192 mg of an isotope is given by A(t)= 192e-0.015t, where t is time in years. Find the amount left after 55 years.

Round your answer to the nearest whole number.
A 83

B 84

C 42

D 189
Mathematics
1 answer:
statuscvo [17]1 year ago
6 0

Answer:

This is a typical radioactive decay problem which uses the general form:

A = A0e^(-kt)

So, in the given equation, A0 = 192 and k = 0.015. We are to find the amount of substance left after t = 55 years. That would be represented by A. The solution is as follows:

A = 192e^(-0.015*55)

A = 84 mg

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During an experiment, Juan rolled a six-sided number cube 18 times. The number two occurred four times. Juan claimed the experim
Maru [420]

Answer:

1) Juans claim is incorrect. The correct experimental probablilty is 2/9

Step-by-step explanation:

8 0
1 year ago
Read 3 more answers
answer A radio station located 120 miles due east of Collinsville has a listening radius of 100 miles. A straight road joins Col
melamori03 [73]

Answer:

168.7602 miles

Step-by-step explanation:

One way to solve this problem is by using an equation that describes the listening radius of the station, and another for the road, then the points where this two-equation intersect each other will represent when the driver starts and stops listening to the station, and the distance between the points is the miles that the driver will receive the signal.

The equation for the listening radius (the radio station is at (0,0)):

x^2+y^2=100^2

The equation for the road that past through the points (-120,0) and (80,100) (Collinsville and Harmony respectively):

m=\frac{y_2-y_1}{x_2-x_1} =\frac{100-0}{80-(-120)}=\frac{100}{200}=\frac{1}{2}

y-y_1=m(x-x_1)\\y-0=\frac{1}{2}(x-(-120))\\ y=\frac{1}{2}x+60

Substitutes the value of y in the equation of the circle:

x^2+(\frac{1}{2}x+60)^2=100^2\\x^2+\frac{1}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600-10000=0\\\frac{5}{4} x^2+60x-6400=0\\5 x^2+240x-25600=0\\x^2+48x-5120=0\\

The formula to solve second-degree equations:

x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a} \\x_{1,2}=\frac{-48\pm\sqrt{48^2-4(1)(-5120)} }{2(1)}\\x_{1,2}=\frac{-48\pm\sqrt{2304+20480} }{2}\\x_{1,2}=\frac{-48\pm\sqrt{22784} }{2}\\x_{1,2}=\frac{-48\pm16\sqrt{89} }{2}\\x_{1,2}=-24\pm8\sqrt{89} \\x_1=-24+8\sqrt{89}\approx51.4718\\x_2=-24-8\sqrt{89}\approx-99.4718\\

Using the values in x to find the values in y:

y_1=\frac{1}{2}x_1+60\\y_1=\frac{1}{2}(-24+8\sqrt{89} )+60\\y_1=-12+4\sqrt{89}+60\\ y_1=48+4\sqrt{89}\approx85.7359

y_2=\frac{1}{2}x_2+60\\y_2=\frac{1}{2}(-24-8\sqrt{89} )+60\\y_1=-12-4\sqrt{89}+60\\ y_1=48-4\sqrt{89}\approx10.2641

The distance between the points (51.4718,85.7359) and (-99.4718,10.2641) :

d=\sqrt{(x_1 -x_2 )^2+(y_1 -y_2)^2} \\d=\sqrt{(-24+8\sqrt{89} -(-24-8\sqrt{89}) )^2+(48+4\sqrt{89} -(48-4\sqrt{89}) )^2}\\d=\sqrt{(-24+8\sqrt{89} +24+8\sqrt{89} )^2+(48+4\sqrt{89} -48+4\sqrt{89} )^2}\\d=\sqrt{(16\sqrt{89} )^2+(8\sqrt{89} )^2}\\d=\sqrt{22784+5696}\\d=\sqrt{28480}\\d=8\sqrt{445}\approx168.7602miles

4 0
1 year ago
For the following right triangle, find the side length x. Round your answer to the nearest hundredth.
tia_tia [17]
<h3>Answer:  6.93</h3>

Work Shown:

Use the pythagorean theorem to find x

a^2 + b^2 = c^2

x^2 + 11^2 = 13^2

x^2 + 121 = 169

x^2 = 169 - 121

x^2 = 48

x = sqrt(48)

x = 6.92820323027551

x = 6.93

5 0
1 year ago
Read 2 more answers
2 cities have nearly the same north-south line 90° W. The latitude of the first city is 23°N, the latitude of the second city is
Ivan

Answer:

The cities are approximately 1452 km apart

Step-by-step explanation:

The two cities are on the same Longitude 90°W therefore the distance between them is on a great circle

Latitude of the first city = 23°N

Latitude of the second city = 36° N

Because their latitude are on the same polar axis, we subtract to get the angular difference

Angular Difference = 36 -23 =13°

Distance= \frac{\alpha}{360} X 2\pi R

where \alpha = Angular Difference, R= radius of the earth

Distance= \frac{13}{360} X 2\pi X 6400=1452.11km

The cities are approximately 1452 km apart

4 0
2 years ago
A batch of 445 containers for frozen orange juice contains 3 that are defective. Two are selected, at random, without replacemen
Elan Coil [88]

Answer:

  1. When Two containers are selected

(a) Probability that the second one selected is defective given that the first one was defective = 0.00450

(b) Probability that both are defective = 0.0112461

(c) Probability that both are acceptable = 0.986

    2. When Three containers are selected

(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451

(c) Probability that all three are defective = 6.855 x 10^{-8} .  

Step-by-step explanation:

We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.

                  Total containers = 445

                   Defective ones   = 3

           Non - Defective ones = 442 { Acceptable ones}

  • Two containers are selected, at random, without replacement from the batch.

(a) Probability that the second one selected is defective given that the first one was defective is given by;

  <em>Since we had selected one defective so for selecting second the available </em>

<em>   containers are 444 and available defective ones are 2 because once </em>

<em>    chosen they are not replaced.</em>

Hence, Probability that the second one selected is defective given that the first one was defective = \frac{2}{444} = 0.00450

(b) Probability that both are defective = P(first being defective) +

                                                                     P(Second being defective)

                 = \frac{3}{445} + \frac{2}{444} = 0.0112461

(c) Probability that both are acceptable = P(First acceptable) +  P(Second acceptable)

Since, total number of acceptable containers are 442 and total containers are 445.

 So, Required Probability = \frac{442}{445}*\frac{441}{444} = 0.986

  • Three containers are selected, at random, without replacement from the batch.

(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;

<em>Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .</em>

Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = \frac{1}{443} = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;

<em>Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.</em>

Hence, Required probability = \frac{2}{443} = 0.00451 .

(c) Probability that all three are defective = P(First being defective) +

                              P(Second being defective) +  P(Third being defective)

        = \frac{3}{445}* \frac{2}{444}  * \frac{1}{443} = 6.855 x 10^{-8} .                

               

5 0
2 years ago
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