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2 years ago

13

What is the answer to -56/6

1 answer:

mafiozo [28]2 years ago

5 0

Answer: -9 1/3 or -9.3333333

Step-by-step explanation: You would just have to divide those two numbers and then add the negative to your answer.

You might be interested in

Find two numbers if their ratio is 9:11 and their difference is 6.

erastova [34]

Answer:

Step-by-step explanation:

27,33

9*3 is 27

11*3 is 33

33-27 is 6

i only have one, sorry!

8 0

2 years ago

Read 2 more answers

A circle has a 10 centimeter radius. Find the area of the circle to the nearest tenth. (Use 3.14 for ?)

pickupchik [31]

Answer:

314.2 cm^2

Step-by-step explanation:

The formula for the area of a circle of radius r is A = 3.14r^2.

With r = 10 cm, r^2 = 100 cm^2, and so the area of this circle is 314.2 cm^2, to the nearest tenth.

3 0

2 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

A pharmacist receives a shipment of 22 bottles of a drug and has 3 of the bottles tested. If 5 of the 22 bottles are contaminate

Norma-Jean [14]

Answer:

There is a 44.16% probability that exactly 1 of the tested bottles is contaminated.

Step-by-step explanation:

$C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem, we have that:

Total number of combinations:

$C_{22,3} = \frac{22!}{3!(18)!} = 1540$

Desired combinations:

It is 1 one 5(contamined) and 2 of 17(non contamined). So:

$C_{5,1}*C_{17,2} = 5*17*8 = 680$

What is the probability that exactly 1 of the tested bottles is contaminated?

$P = \frac{680}{1540} = 0.4416$

There is a 44.16% probability that exactly 1 of the tested bottles is contaminated.

5 0

3 years ago

11 / 2 plus 11 / 4 in fraction

dybincka [34]

Answer:

8and1/4

Step-by-step explanation:

8 0

2 years ago