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2 years ago

What is a reasonable percent error in chemistry?

1 answer:

8 0

For physics, we use half the smallest division of the equipment used.
For chemistry, it should be the same.

The formula for percent error: delta(x)/x * 100% = delta(y)/y

Hope I helped :)

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Jason agreed to make 12 payments of $38.50 to buy a new T.V. SO far he has made 8 payment. How much more money does he owe?

alexira [117]

Answer:

154$

Step-by-step explanation:

8 0

2 years ago

15 + r as a verbal expression

kodGreya [7K]

Hey there!

<h2>ANSWER: $r+15$ </h2><h2>EXPLANATION:</h2><h2> $15+r$ </h2>

Simplify by flipping.

$r+15$

Hope this helps!

- TestedHyperr

7 0

2 years ago

Find the slope of the line.<br><br><br> please help<br><br><br> will mark brainlyest

Mamont248 [21]

Answer:

m = 1

Step-by-step explanation:

slope = (y₂-y₁)/(x₂-x₁)

m = (4 - -1) / 1 - -4) = 5/5 = 1

5 0

2 years ago

The Statistics Club at Woodvale College sold college Y-shirts as a fundraiser. The result of the sale are shown below. Choose on

zavuch27 [327]

We are asked to find the probability that the student sold 11-15 T-shirts or less than 6 T-shirts.

The total number of club members = 2 + 14 + 12 + 3 + 6 + 1 = 38

Probability of 11-15 T-shirts:

From the table, 3 club members bought 11 - 15 T-shirts.

So the probability is 3/38

Probability of less than 6 T-shirts:

Less than 6 T-shirts means 1 - 5 plus 0 T-shirts.

Form the table, 14 club members bought 1 - 5 T-shirts and 2 club members bought 0 T-shirts.

Total club member who bought 1 - 5 and 0 T-shirts = 14 + 2 = 16

So the probability is 16/38

Now we have to add them together to find the probability that the student sold 11-15 T-shirts or less than 6 T-shirts.

(or means to add)

So the probability is 3/38 + 16/38 = 19/38 = 0.5000

Therefore, the correct answer is option c.

6 0

4 months ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

2 years ago