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2 years ago

6

Helppp plzzzzzzzzzzz

1 answer:

Sphinxa [80]2 years ago

4 0

The answer is .5 boxes assembled per minute

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Element X decays radioactively with a half life of 12 minutes. If there are 200 grams of Element X, how long, to the nearest ten

Semenov [28]

Answer:

It would take 24 minutes for the element to decay to 50 grams

Step-by-step explanation:

The equation for the amount of the element present, after t minutes, is:

$Q(t) = Q(0)e^{-rt}$

In which Q(X) decays radioactively with a half life of 12 minutes.(0) is the initial amount and r is the rate it decreases.

Half life of 12 minutes

This means that $Q(12) = 0.5Q(0)$

So

$Q(t) = Q(0)e^{-rt}$

$0.5Q(0) = Q(0)e^{-12r}$

$e^{-12r} = 0.5$

$\ln{e^{-12r}} = \ln{0.5}$

$-12r = \ln{0.5}$

$12r = -\ln{0.5}$

$r = -\frac{\ln{0.5}}{12}$

$r = 0.05776$

If there are 200 grams of Element X, how long, to the nearest tenth of a minute, would it take the element to decay to 50 grams?

This is t when Q(t) = 50. Q(0) = 200.

$Q(t) = Q(0)e^{-rt}$

$50 = 200e^{-0.05776t}$

$e^{-0.05776t} = 0.25$

$\ln{e^{-0.05776t}} = \ln{0.25}$

$-0.05776t = \ln{0.25}$

$0.05776t = -\ln{0.25}$

$t = -\frac{\ln{0.25}}{0.05776}$

$t = 24$

It would take 24 minutes for the element to decay to 50 grams

6 0

3 years ago

One satellite is scheduled to be launched from Cape Canaveral in Florida, and another launching is scheduled for Vandenberg Air

Lisa [10]

Answer:

$P(A) = 0.45$

$P(B) = 0.32$

Step-by-step explanation:

Given

$P(A) > P(B)$

$P(A\ u\ B) = 0.626$

$P(A\ n\ B) = 0.144$

Required

Find P(A) and P(B)

We have that:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$ --- (1)

and

$P(A\ n\ B) = P(A) * P(B)$ --- (2)

The equations become:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$ --- (1)

$0.626 = P(A) + P(B) - 0.144$

Collect like terms

$P(A) + P(B) = 0.626 + 0.144$

$P(A) + P(B) = 0.770$

Make P(A) the subject

$P(A) = 0.770 - P(B)$

$P(A\ n\ B) = P(A) * P(B)$ --- (2)

$0.144 = P(A) * P(B)$

$P(A) * P(B) = 0.144$

Substitute: $P(A) = 0.770 - P(B)$

$[0.770 - P(B)] * P(B) = 0.144$

Open bracket

$0.770P(B) - P(B)^2 = 0.144$

Represent P(B) with x

$0.770x - x^2 = 0.144$

Rewrite as:

$x^2 - 0.770x + 0.144 = 0$

Expand

$x^2 - 0.45x - 0.32x + 0.144 = 0$

Factorize:

$x[x - 0.45] - 0.32[x - 0.45]= 0$

Factor out x - 0.45

$[x - 0.32][x - 0.45]= 0$

Split

$x - 0.32= 0 \ or\ x - 0.45= 0$

Solve for x

$x = 0.32\ or\ x = 0.45$

Recall that:

$P(B) = x$

So, we have:

$P(B) = 0.32 \ or \ P(B) = 0.45$

Recall that:

$P(A) = 0.770 - P(B)$

So, we have:

$P(A) = 0.770 - 0.32 \ or\ P(A) =0.770 - 0.45$

$P(A) = 0.45 \ or\ P(A) =0.32$

Since:

$P(A) > P(B)$

Then:

$P(A) = 0.45$

$P(B) = 0.32$

6 0

2 years ago

Please help me...I need to turn this in PLZZZZ!!!!!!

Nuetrik [128]

Answer:

i believe its C

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Luis measures a plant and finds it is 12.5 centimeters tall.

VARVARA [1.3K]

Answer:

125mm

Step-by-step explanation:

12.5 x 10 = 125

6 0

2 years ago

Read 2 more answers

If you start at vertex A and use the "shortest route" algorithm, what would be the second path to

MAVERICK [17]

Answer:

Shortest distance=|AC|/ $\sqrt[2]{2}$ .

Kindly find the attached for the figure

Step-by-step explanation:

This problem can be addressed using right-angled,

Let have right angle triangle with the shortest distance=hypotenuse;

hypotenuse=|AC|

using pythagoras theorem, we have

|AC|^2=|AB|^2+|BC|^2

Let |AB|=|BC|

|AC|^2=2|AB|^2

|AB|=|AC|/ $\sqrt[2]{2}$

Shortest distance=|AC|/ $\sqrt[2]{2}$

7 0

2 years ago