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Irina-Kira [14]
3 years ago
15

Question 1

Mathematics
1 answer:
inessss [21]3 years ago
4 0

Answer:

Least possible value of b is 9.

Step-by-step explanation:

It is given that 7b-8>4b+16 and b is an odd integer.

We need to find the least possible value of b.

We have,

7b-8>4b+16

Isolate variable terms.

7b-4b>8+16

3b>24

Divide both sides by 3.

b>8

Since b>8 and b an odd integer, therefore the possible values of b are 9, 11, 13, 15, ... .

Hence, the least possible value of b is 9.

You might be interested in
Modeling Radioactive Decay In Exercise, complete the table for each radioactive isotope.
Julli [10]

Answer:

Step-by-step explanation:

Hello!

The complete table attached.

The following model allows you to predict the decade rate of a substance in a given period of time, i.e. the decomposition rate of a radioactive isotope is proportional to the initial amount of it given in a determined time:

y= C e^{kt}

Where:

y represents the amount of substance remaining after a determined period of time (t)

C is the initial amount of substance

k is the decaing constant

t is the amount of time (years)

In order to know the decay rate of a given radioactive substance you need to know it's half-life. Rembember, tha half-life of a radioactive isotope is the time it takes to reduce its mass to half its size, for example if you were yo have 2gr of a radioactive isotope, its half-life will be the time it takes for those to grams to reduce to 1 gram.

1)

For the first element you have the the following information:

²²⁶Ra (Radium)

Half-life 1599 years

Initial quantity 8 grams

Since we don't have the constant of decay (k) I'm going to calculate it using a initial quantity of one gram. We know that after 1599 years the initial gram of Ra will be reduced to 0.5 grams, using this information and the model:

y= C e^{kt}

0.5= 1 e^{k(1599)}

0.5= e^{k(1599)}

ln 0.5= k(1599)

\frac{1}{1599} ln 0.05 = k

k= -0.0004335

If the initial amount is C= 8 grams then after t=1599 you will have 4 grams:

y= C e^{kt}

y= 8 e^{(-0.0004355*1599)}

y= 4 grams

Now that we have the value of k for Radium we can calculate the remaining amount at t=1000 and t= 10000

t=1000

y= C e^{kt}

y_{t=1000}= 8 e^{(-0.0004355*1000)}

y_{t=1000}= 5.186 grams

t= 10000

y= C e^{kt}

y_{t=10000}= 8 e^{(-0.0004355*10000)}

y_{t=10000}= 0.103 gram

As you can see after 1000 years more of the initial quantity is left but after 10000 it is almost gone.

2)

¹⁴C (Carbon)

Half-life 5715

Initial quantity 5 grams

As before, the constant k is unknown so the first step is to calculate it using the data of the hald life with C= 1 gram

y= C e^{kt}

1/2= e^{k5715}

ln 1/2= k5715

\frac{1}{5715} ln1/2= k

k= -0.0001213

Now we can calculate the remaining mass of carbon after t= 1000 and t= 10000

t=1000

y= C e^{kt}

y_{t=1000}= 5 e^{(-0.0001213*1000)}

y_{t=1000}= 4.429 grams

t= 10000

y= C e^{kt}

y_{t=10000}= 5 e^{(-0.0001213*10000)}

y_{t=10000}= 1.487 grams

3)

This excersice is for the same element as 2)

¹⁴C (Carbon)

Half-life 5715

y_{t=10000}= 6 grams

But instead of the initial quantity, we have the data of the remaining mass after t= 10000 years. Since the half-life for this isotope is the same as before, we already know the value of the constant and can calculate the initial quantity C

y_{t=10000}= C e^{kt}

6= C e^{(-0.0001213*10000)}

C= \frac{6}{e^(-0.0001213*10000)}

C= 20.18 grams

Now we can calculate the remaining mass at t=1000

y_{t=1000}= 20.18 e^{(-0.0001213*1000)}

y_{t=1000}= 17.87 grams

4)

For this exercise we have the same element as in 1) so we already know the value of k and can calculate the initial quantity and the remaining mass at t= 10000

²²⁶Ra (Radium)

Half-life 1599 years

From 1) k= -0.0004335

y_{t=1000}= 0.7 gram

y_{t=1000}= C e^{kt}

0.7= C e^{(-0.0004335*1000)}

C= \frac{0.7}{e^(-0.0004335*1000)}

C= 1.0798 grams ≅ 1.08 grams

Now we can calculate the remaining mass at t=10000

y_{t=10000}= 1.08 e^{(-0.0001213*10000)}

y_{t=10000}= 0.32 gram

5)

The element is

²³⁹Pu (Plutonium)

Half-life 24100 years

Amount after 1000 y_{t=1000}= 2.4 grams

First step is to find out the decay constant (k) for ²³⁹Pu, as before I'll use an initial quantity of C= 1 gram and the half life of the element:

y= C e^{kt}

1/2= e^{k24100}

ln 1/2= k*24100

k= \frac{1}{24100} * ln 1/2

k= -0.00002876

Now we calculate the initial quantity using the given information

y_{t=1000}= C e^{kt}

2.4= C e^{( -0.00002876*1000)}

C= \frac{2.4}{e^( -0.00002876*1000)}

C=2.47 grams

And the remaining mass at t= 10000 is:

y_{t=10000}= C e^{kt}

y_{t=10000}= 2.47 * e^{( -0.00002876*10000)}

y_{t=10000}= 1.85 grams

6)

²³⁹Pu (Plutonium)

Half-life 24100 years

Amount after 10000 y_{t=10000}= 7.1 grams

From 5) k= -0.00002876

The initial quantity is:

y_{t=1000}= C e^{kt}

7.1= C e^{( -0.00002876*10000)}

C= \frac{7.1}{e^( -0.00002876*10000)}

C= 9.47 grams

And the remaining masss for t=1000 is:

y_{t=1000}= C e^{kt}

y_{t=1000}= 9.47 * e^{( -0.00002876*1000)}

y_{t=1000}= 9.20 grams

I hope it helps!

4 0
3 years ago
( 8 2/4 + 2 1/6) - 5 2/12
Readme [11.4K]

Answer:

5 1/2 or 5.5

Step-by-step explanation:

=8 1/2+2 1/6-5 2/12

=17/2+ 2 1/6-5 2/12

=17/2+13/6-5 2/12

=32/3- 5 2/12

=32/3-31/6

=11/2

=5 1/2

5 0
2 years ago
In a poll taken in December 2012, Gallup asked 1006 national adults whether they were baseball fans: 48% said they were. Almost
Yuki888 [10]

Answer:

0.0259,0.0406

Step-by-step explanation:

Given that in a poll taken in December 2012, Gallup asked 1006 national adults whether they were baseball fans: 48% said they were. Almost five years earlier, in February 2008, only 35% of a similar-size sample had reported being baseball fans.

a) p is the proportion of national adults who are baseball fans.

b) For 2012 poll, p=0.48

Std error = \sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.48*0.52}{1006} } \\=0.0158

Margin of error at 90% = 1.645 * std error

= 0.0259

c) Margin of error is the admissible limit on either side of the mean where the sample mean can lie

d) Margin of error would be larger for 99% because critical value 2.58 >1.645

e) Margin of error 99% = 2.58*0.0157\\=0.0406

6 0
3 years ago
How many centiliters are in 6.02 decaliters
Vinil7 [7]
Well one decaliter is 1000 centiliters so 6.02 * 1000 = 6020 . so the answer is 6,020 centiliters .
8 0
3 years ago
If you measure 5/6 meter how many twelfths have you measured
Alinara [238K]
1/6 = 2/12 because 12 is 2 times 6.
Multiply both numbers by 2 to get your answer:
5/6 = 10/12
4 0
3 years ago
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