A piece of dry ice is in the shape of a cube with edge lengths of 7 centimeters. Find the surface area of the dry ice

The half-life of a certain radioactive substance is 45 days. There are 6.2 grams present initially. On what day

kvasek [131]

Answer:

There will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days

Step-by-step explanation:

The given parameters are;

The half life of the radioactive substance = 45 days

The mass of the substance initially present = 6.2 grams

The expression for evaluating the half life is given as follows;

$N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}}$

Where;

N(t) = The amount of the substance left after a given time period = 1 gram

N₀ = The initial amount of the radioactive substance = 6.2 grams

$t_{1/2}$ = The half life of the radioactive substance = 45 days

Substituting the values gives;

$1 = 6.2 \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}$

$\dfrac{1}{6.2} = \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}$

$ln\left (\dfrac{1}{6.2} \right ) = {\dfrac{t}{45} \times ln \left (\dfrac{1}{2} \right )$

$t = 45 \times \dfrac{ln\left (\dfrac{1}{6.2} \right ) }{ln \left (\dfrac{1}{2} \right )} \approx 118.45 \ days$

The time that it takes for the mass of the radioactive substance to remain 1 g ≈ 118.45 days

Therefore, there will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days.

3 0

3 years ago

Two urns both contain green balls and red balls. Urn I contains 6 green balls and 4 red balls and Urn II contains 8 green balls

34kurt

Presumably, whatever is drawn from Urn I is independent of what is drawn from Urn II. This means

$P(\text{red from Urn 1 AND red from Urn 2})=P(\text{red from Urn 1})\cdot P(\text{red from Urn 2})$

We have

$P(\text{red from Urn 1})=\dfrac4{10}=\dfrac25$

$P(\text{red from Urn 1})=\dfrac7{15}$

so the probability of drawing red balls from both urns is $\dfrac25\cdot\dfrac7{15}=\dfrac{14}{75}$

3 0

3 years ago

Please help me with steps.

laiz [17]

Answer:

I think its would be d

Step-by-step explanation:

the distributive property

8 0

3 years ago

Read 2 more answers

*please help me out if you know how to solve, last person I asked just put a random awnser for point, the problem is bellow and

kicyunya [14]

For both of these, we will divide the wanted outcome by the number of possible outcomes. We will end up with a decimal. A decimal multiplied by 100 becomes a percent.

[a] 50%

$\displaystyle \frac{\text{wanted outcomes}}{\text{possible outcomes}} =\frac{\text{student failed}}{\text{Period 2's class}} =\frac{12}{24} =0.5,\:0.5*100=50\%$

[b] 52%

$\displaystyle \frac{\text{wanted outcomes}}{\text{possible outcomes}} =\frac{\text{student failed from Period 3}}{\text{fails from both classes}} =\frac{13}{25} =0.52,$

$\displaystyle 0.52*100=52\%$

8 0

2 years ago

Plssssss help due today

Grace [21]

10^2 - 8^2
= 100 - 64
= 36
LM = √36 = 6

answer
LM = 6

8 0

3 years ago