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2 years ago

3. Marcus's family has completed 30% of a trip. They have traveled 21 miles.

Mathematics

1 answer:

vlada-n [284]2 years ago

3 0

Answer: 70 miles

Step-by-step explanation: scince the trip is 30% finished and that 30% was 21 miles you just have to multiply 21 by 3 which is 63and than you have to figure out what 1 third of 21 is which is 7 and than you just have to add 7 to 63. Hope this helps

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A bag contains 6 green counters, 4 blue counters and 2 red counters. Two counters are drawn from the bag at random without repla

Svetach [21]

Answer:

24.24%

Step-by-step explanation:

In other words we need to find the probability of getting one blue counter and another non-blue counter in the two picks. Based on the stats provided, there are a total of 12 counters (6 + 4 + 2), out of which only 4 are blue. This means that the probability for the first counter chosen being blue is 4/12

Since we do not replace the counter, we now have a total of 11 counters. Since the second counter cannot be blue, then we have 8 possible choices. This means that the probability of the second counter not being blue is 8/11. Now we need to multiply these two probabilities together to calculate the probability of choosing only one blue counter and one non-blue counter in two picks.

$\frac{4}{12} * \frac{8}{11} = \frac{32}{132}$ or 0.2424 or 24.24%

8 0

3 years ago

Paula used these steps to solve an equation:

Nata [24]

Answer: Steps 4 and 5

Step-by-step explanation:

6 0

3 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

Write 0.36 as a fraction in simplest form

Vsevolod [243]

Answer:

9/25

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Two ships leave a harbor at the same time. One ship travels on a bearing Upper S 14 degrees Upper W at 15 miles per hour. The ot

jekas [21]

Answer:

69.94 miles

Step-by-step explanation:

The distance d₁ the first ship moves after 3 hours is 3 hours × 15 miles per hours = 45 miles

The distance d₂ the second ship moves after 3 hours is 3 hours × 12 miles per hours = 36 miles.

The angle the first ship's direction makes in the North-East direction is 90° - 75° = 15°

The angle the second ship's direction makes in the South-West direction = 14°

The distance moved by the two ships form the side of a triangle. The angle, θ between the two ship directions is 14° + 90° + 15° = 119°

Using the cosine rule, we find the distance d between the two ships

d = √(d₁² + d₂² -2d₁d₂cosθ)

= √(45² + 36² -2×45×36cos119°)

= √(2025 + 1296- (-1570.78))

= √(3321 + 1570.78)

= √4891.78

= 69.94 miles

4 0

3 years ago