Ava-May Nelson
2021-03-07
Answered

Lodine - 131, a radioactive substance that is effective in localing brain tumors, a half-life of only eight days. A hospital purchased 22 grams of the substance but had to wail five days before it could be used. How much of the substance was left after five days?

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Tuthornt

Answered 2021-03-08
Author has **107** answers

Data analysis

Given data,

Initial amount of substance

Half-life

Time period

Amount of substance left after 't' = ?

Solution

The decay of an radioactive substance is exponential.

And the amount of substance (P) left fter 't' days is given by,

Where lambda is decay constant.

Given

That is by 8 days, amount left will become half.

Now after 5 days, amount left,

Hence amount left after 5 days is 14.265 grams (rounded to 3 decimals)

asked 2020-12-01

According to a study by Dr. John McDougall of his live-in weight loss program at St. Helena Hospital, the people who follow his program lose between 6 and 15 pounds a month until they approach trim body weight. Let's suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month. Give the distribution of X. Enter an exact number as an integer, fraction, or decimal.

asked 2021-01-06

The product of 2 decimals is 20.062 one of the factors has 2 decimals .how many decimals in other factors.

asked 2021-05-11

Bethany needs to borrow $\$10,000.$ She can borrow the money at $5.5\mathrm{\%}$ simple interest for 4 yr or she can borrow at $5\mathrm{\%}$ with interest compounded continuously for 4 yr.

a) How much total interest would Bethany pay at$5.5\mathrm{\%}$ simple interest?

b) How much total interest would Bethany pay at$5$ interest compounded continuously?

c) Which option results in less total interest?

a) How much total interest would Bethany pay at

b) How much total interest would Bethany pay at

c) Which option results in less total interest?

asked 2020-10-21

On average, 3 traffic accidents per month occur at a certain intersection. What is the probability
that in any given month at this intersection

(a) exactly 5 accidents will occur?

(b) fewer than 3 accidents will occur?

(c) at least 2 accidents will occur?

(a) exactly 5 accidents will occur?

(b) fewer than 3 accidents will occur?

(c) at least 2 accidents will occur?

asked 2022-06-13

Help with simplification of a rational expression (with fractional powers)

Can you please help me see what I don't see yet. Here's a problem from a high school textbook (ISBN 978-5-488-02046-7 p.9, #1.029):

$\frac{({a}^{1/m}-{a}^{1/n}{)}^{2}\cdot 4{a}^{(m+n)/mn}}{({a}^{2/m}-{a}^{2/n})(\sqrt[m]{{a}^{m+1}}+\sqrt[n]{{a}^{n+1}})}$

Here's my try at it:

$\frac{({a}^{1/m}-{a}^{1/n})({a}^{1/m}-{a}^{1/n})\cdot 4{a}^{(1/m)+(1/n)}}{({a}^{1/m}-{a}^{1/n})({a}^{1/m}+{a}^{1/n})\cdot a({a}^{1/m}+{a}^{1/n})}$

...which is

$\frac{({a}^{1/m}-{a}^{1/n})\cdot 4{a}^{(1/m)+(1/n)}}{a({a}^{1/m}+{a}^{1/n}{)}^{2}}$

Wolfram Alpha's simplify stops here, too. I don't see where to go from here. The final form, according to the book, is this:

$\frac{1}{a({a}^{1/m}-{a}^{1/n})}$

How did they do it?

Can you please help me see what I don't see yet. Here's a problem from a high school textbook (ISBN 978-5-488-02046-7 p.9, #1.029):

$\frac{({a}^{1/m}-{a}^{1/n}{)}^{2}\cdot 4{a}^{(m+n)/mn}}{({a}^{2/m}-{a}^{2/n})(\sqrt[m]{{a}^{m+1}}+\sqrt[n]{{a}^{n+1}})}$

Here's my try at it:

$\frac{({a}^{1/m}-{a}^{1/n})({a}^{1/m}-{a}^{1/n})\cdot 4{a}^{(1/m)+(1/n)}}{({a}^{1/m}-{a}^{1/n})({a}^{1/m}+{a}^{1/n})\cdot a({a}^{1/m}+{a}^{1/n})}$

...which is

$\frac{({a}^{1/m}-{a}^{1/n})\cdot 4{a}^{(1/m)+(1/n)}}{a({a}^{1/m}+{a}^{1/n}{)}^{2}}$

Wolfram Alpha's simplify stops here, too. I don't see where to go from here. The final form, according to the book, is this:

$\frac{1}{a({a}^{1/m}-{a}^{1/n})}$

How did they do it?

asked 2022-06-20

Let $(X,\mathcal{A},\nu )$ be a measure space. Suppose that the measure of $X$ is bounded. Show that for disjoint ${A}_{1},{A}_{2},\dots $ $\underset{n\to \mathrm{\infty}}{lim}\nu ({A}_{n})=0$.

Since $X$ is bounded from countable additivity we have that

$\nu \left(\bigcup _{n=1}^{\mathrm{\infty}}{A}_{n}\right)=\sum _{n=1}^{\mathrm{\infty}}\nu ({A}_{n})\le \nu (X)<\mathrm{\infty}.$

Thus from Borel-Cantelli

$\sum _{n=1}^{\mathrm{\infty}}\nu ({A}_{n})<\mathrm{\infty}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\nu (\underset{n\to \mathrm{\infty}}{lim\u2006sup}{A}_{n})=0.$

Now I also have that

$\nu (\underset{n\to \mathrm{\infty}}{lim\u2006sup}{A}_{n})\ge \underset{n\to \mathrm{\infty}}{lim\u2006sup}\nu ({A}_{n})$

which implies that

$\underset{n\to \mathrm{\infty}}{lim\u2006sup}\nu ({A}_{n})=0.$

How can I deduce from here that $\underset{n\to \mathrm{\infty}}{lim}\nu ({A}_{n})=0$? It seems I would need to use the fact that

$0=\underset{n\to \mathrm{\infty}}{lim\u2006sup}\nu ({A}_{n})\ge \underset{n\to \mathrm{\infty}}{lim\u2006inf}\nu ({A}_{n})\ge 0$

so $\underset{n\to \mathrm{\infty}}{lim\u2006inf}\nu ({A}_{n})=0$ also and this would imply that $\underset{n\to \mathrm{\infty}}{lim}\nu ({A}_{n})=0$ because of?

Since $X$ is bounded from countable additivity we have that

$\nu \left(\bigcup _{n=1}^{\mathrm{\infty}}{A}_{n}\right)=\sum _{n=1}^{\mathrm{\infty}}\nu ({A}_{n})\le \nu (X)<\mathrm{\infty}.$

Thus from Borel-Cantelli

$\sum _{n=1}^{\mathrm{\infty}}\nu ({A}_{n})<\mathrm{\infty}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\nu (\underset{n\to \mathrm{\infty}}{lim\u2006sup}{A}_{n})=0.$

Now I also have that

$\nu (\underset{n\to \mathrm{\infty}}{lim\u2006sup}{A}_{n})\ge \underset{n\to \mathrm{\infty}}{lim\u2006sup}\nu ({A}_{n})$

which implies that

$\underset{n\to \mathrm{\infty}}{lim\u2006sup}\nu ({A}_{n})=0.$

How can I deduce from here that $\underset{n\to \mathrm{\infty}}{lim}\nu ({A}_{n})=0$? It seems I would need to use the fact that

$0=\underset{n\to \mathrm{\infty}}{lim\u2006sup}\nu ({A}_{n})\ge \underset{n\to \mathrm{\infty}}{lim\u2006inf}\nu ({A}_{n})\ge 0$

so $\underset{n\to \mathrm{\infty}}{lim\u2006inf}\nu ({A}_{n})=0$ also and this would imply that $\underset{n\to \mathrm{\infty}}{lim}\nu ({A}_{n})=0$ because of?

asked 2022-04-21

Multiply

$(-6x-5)({x}^{3}-x+2)$