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4 years ago

Write the phrase as an expression. Then evaluate when x=5. The sum of a number x and 4, all divided by 3

2 answers:

3 0

Sum of a number, x and 4 all divided by 3

sum of x and 4 means x+4
all divided by 3
$\frac{x+4}{3}$
evaluate for x=5

$\frac{5+4}{3}=$
$\frac{9}{3}=$
$3$

it's 3

lisabon 2012 [21]4 years ago

3 0

Answer:

STOP

Step-by-step explanation:

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What fraction is closet to 0.195 <br><br> 1/2. 1/4. 1/8 1/10

erastovalidia [21]

1/4. i believe but im not too sure

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3 years ago

Find integers a, b, and c such that a) a | (b - c) but a - b and a - c b) a | (2b + 3c) but a - b and a - c

seraphim [82]

Answer:

Part a) (a,b,c)=(10,19,9)

Part b) (a,b,c)=(9,3,1)

Please let be know if I have interpreted your question correctly. Thank you.

Step-by-step explanation:

Part a)

This is the way I interpret part a:

Find integers a,b,c such that a|(b-c) but a does not divide b and a does not divide c.

There are many triples satisfying part a.

One such triple is (a,b,c)=(3,10,7).

3 divides 10-7

But 3 does not divide 10 and 3 does not divide 7.

Another example is (a,b,c)=(2,7,5).

2 divides 7-5

But 2 doesn't divide 7 and 2 doesn't divide 5.

I will give another example.

(a,b,c)=(10,19,9)

10 divides 19-9

But 10 doesn't divide 19 and 10 doesn't divide 9

Part b)

This is the say I interpret part b:

a | (2b + 3c) but a does not divide b and a does not divide c

So lets let b=2 and c=5, then 2b+3c=19.

We just need to choose an a such that it divides 19 but not 2 and 5. That should be easy. 19 itself will do that.

So one (a,b,c) could be (19,2,5).

Let's try for another example.

Let b=3 and c=1.

Then 2b+3c=9.

So we just need to find an a that divides 9 but not 3 and not 1.

9 works.

So another example is (9, 3,1).

5 0

3 years ago

For each carbon-14 atom, there are approximately 10^12 carbon-12 atoms. This ratio is constant in each organism. The carbon-12 r

Bond [772]

Answer:

1) There are 100 carbon-14 atoms in the wood when it died

2) A function that models the number 'N' of carbon-14 atoms present in the wood (t) years after it died is presented as follows;

$N = 100 \times \left (\dfrac{1}{2} \right )^{\dfrac{t}{5,730} }$

Step-by-step explanation:

The given parameters are;

The number of carbon-12 for each carbon-14 atom ≈ 10¹² carbon-12 atoms

The number of carbon-12 in the piece of wood = 10¹⁴ carbon-12 atoms

The number of carbon-14 in the piece of wood = 40 carbon-14 atoms

The number of carbon-12 in an organism = Constant

The number of carbon-14 in an organism = Decays

1) Given that the number of carbon-12 in an organism is constant, and there are 10¹² carbon-12 atoms per each carbon-14 atom, therefore, we have;

The number of carbon-12 atoms in the wood when it died = 10¹⁴ carbon-12 atoms

The number of carbon-14 atoms in the wood when it died = (10¹⁴ carbon-12 atoms)/(10¹² carbon-12 atoms/(carbon-14 atom)) = 100 carbon-14 atoms

The number of carbon-14 atoms in the wood when it died = 100 carbon-14 atoms

2) The half life of a radioactive isotope is given by the following formula;

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

The half life of carbon-14 atoms, $t_{\frac{1}{2} }$ ≈ 5,730 years

N₀ = The amount of carbon-14 present in the wood when it died = 100 carbon-14 atoms

Therefore, we have;

The number, N, of carbon-14 atoms present in the wood (t) years after it died is given as follows;

$N = 100 \times \left (\dfrac{1}{2} \right )^{\dfrac{t}{5,730} }$

3 0

3 years ago

Let f(x) =7x-13. Find f -1(x)

muminat

If you’re trying to find f(-1) it would be -20

5 0

3 years ago

Read 2 more answers

A survey was conducted that asked 1011 people how many books they had read in the past year. results indicated that x overbarequ

Tcecarenko [31]

Since we do not know the population standard deviation, we will use the t-distribution to construct the 90% confidence interval of the mean number of books people read. The value of $t_{\frac{\alpha }{2}}$ with 1010 degrees of freedom and with alpha equals to 0.10 is 1.64

First, we need to solve for the margin of error, E.
$E=t_{\frac{\alpha }{2}}\cdot \frac{s}{\sqrt{n}}=1.64\cdot \frac{16.6}{\sqrt{1011}}=0.86$

The lower bound of the confidence interval is
$LB=\overline{x}-E=14.8-0.86=13.94$

The upper bound of the confidence interval is
$LB=\overline{x}+E=14.8+0.86=15.66$

Therefore, the 90% confidence interval is (13.94, 15.66).

We are 90% confident that the interval from 13.94 books to 15.66 books does contain the true value of the population mean number of books people read.

4 0

4 years ago