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Elza [17]
3 years ago
14

The probability of waiting 4 minutes or longer for checkout at a particular supermarket counter is 0.1. On a given day, a man an

d his wife decide to shop individually at the market, each checking out at different counters. They both reach the counters at the same time.

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
3 0

Answer:

Step-by-step explanation:

Hope this photo will find you well

Good day!

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Brian pays £465.98 a year on his car insurance.
denpristay [2]

Answer:

<em>£ 449.20</em>

Step-by-step explanation:

We have to reduce 465.98 by 3.6%

Firstly, lets convert the percentage to decimal by dividing by 100. So we have:

3.6/100 = 0.036

Lets find 3.6% of 465.98

That is:

0.036 * 465.98 = 16.77528

This is the "LESSENED" amount. So the final amount would be:

465.98 - 16.77528 = 449.20472

<em>Rounding to 2 decimal places, £ 449.20</em>

8 0
3 years ago
Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot
Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

\frac{dm}{dt} = -\frac{t}{\tau}

Where:

m - Current mass of the isotope, measured in kilograms.

t - Time, measured in days.

\tau - Time constant, measured in days.

The solution of the differential equation is:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }

Where m_{o} is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}

\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }

The half-life of a isotope (t_{1/2}) as a function of time constant is:

t_{1/2} = \tau \cdot \ln2

t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2

The half-life difference between isotope B and isotope A is:

\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|

If \frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9, t_{A} = 33\,days and t_{B} = 43\,days, the difference in the half-lives of the isotopes is:

\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|

\Delta t_{1/2} \approx 65.788\,days

The approximate difference in the half-lives of the isotopes is 66 days.

4 0
3 years ago
Read 2 more answers
What is the surface area of a sphere with radius 4?
mario62 [17]

Answer: 64 units

Step-by-step explanation:

Remember, the formula to find surface area given radius is 4\pi r^2. Let's add in the given information:

4pi4^2 =

4pi16

64pi!

Hope this helps, please consider making me Brainliest.

5 0
1 year ago
Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4
djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0
3 years ago
Read 2 more answers
can someone please help me out with this problem i been trying to figure out what the answer is .. ? or at least give me some ti
k0ka [10]

Answer: sin

Step-by-step explanation:

5 0
3 years ago
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