PLEASE I REALLY NEED HELP

A particular isotope has a half-life of 74 days. If you start with 1 kilogram of this isotope, how much will remain after 150

shtirl [24]

$\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}$

$\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}$

6 0

3 years ago

What is the area of the figure? Use 3.14 for π. Round to the nearest hundredth if necessary.

vfiekz [6]

Answer:

81.12 cm²

Step-by-step explanation:

The area for a semicircle can be figured with this formula: (πr²)/2. The triangle's area formula is bh/2.

(3.14 * 4²)/2 + (14 * 8)/2 (Work the operations in parentheses first.)

(3.14 * 16)/2 + 112/2

50.24/2 + 112/2 (Divide 50.24 by 2. Do the same for 112.)

25.12 + 56 (Add 25.12 & 56.)

81.12 cm²

The area of the figure is 81.12 cm².

7 0

3 years ago

Question 21 of 25

Phantasy [73]

It is A hope this helps :)

5 0

2 years ago

Suppose a random variable, x, arises from a binomial experiment. If n = 6, and p = 0.30, find the following probabilities using

AlladinOne [14]

Answer:

a) P(X=1) = 0.302526

b) P(X=5) = 0.010206

c) P(X=3) = 0.18522

d) P(X≤3) = 0.92953

e) P(X≥5) = 0.010935

f) P(X≤4) = 0.989065

Explanation:

Binomial experiments are modeled by the formula:

$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{(n-x)}$

Where

P(X=x) is the probability of exactly x successes

$C(n,x)=\dfrac{n!}{x!\cdot (n-x)!}$

p is the probability of one success, which must be the same for every trial, and every trial must be independent of other trial.

n is the number of trials

1 - p is the probability of fail

there are only two possible outcomes for each trial: success or fail.

a.) P (x=1)

$P(X=1)=\dfrac{6!}{1!\cdot (6-1)!}\times (0.3)^1\times(1-0.3)^{(6-1)}=0.302526$

b.) P (x=5)

$P(X=5)=\dfrac{5!}{5!\cdot (6-5)!}\times (0.3)^5\times (1-0.3)^{(6-5)}=0.010206$

c.) P (x=3)

Using the same formula:

$P(X=3)=0.18522$

d.) P (x less than or equal to 3)

P(X≤3)= P(X=3) + P(X=2) + P(X=1) + P(X=0)

Also,

P(X≤3) = 1 - P(X≥4) = 1 - P(X=4) - P(X=5) - P(X=6)

You can use either of those approaches. The result is the same.

Using the second one:

P(X=4) = 0.059335
P(X=5) = 0.010206
P(X=6) = 0.000729

P(X≤3) = 1 - 0.05935 - 0.010206 - 0.000729 = 0.92953

e.) P(x greather than or equal to 5)

P(X≥5) = P(X=5) + P(X=6)

P(X≥5) = 0.010206 + 0.000729 = 0.010935

f.) P(x less than or equal 4)

P(X≤4) = 1 - P(X≥5) = 1 - P(X=5) - P(X=6)

P(X≤4) = 1 - 0.010206 - 0.000729 = 0.989065

5 0

3 years ago

Given: ΔABC; a= 10; b = 4, and c = 11. Find the measurement of ∠B to the nearest tenth of a degree.

AleksAgata [21]

Use the Cosine Rule to find this

cos <B = [b^2 - a^2 - c^2] / -2ac

= [ 4^2 - 10^2 - 11^2] / -2*10*11

= 0.93182

so m < B = 21.3 degrees to nearest tenth. Answer

8 0

3 years ago