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3 years ago

9

Alright Soldier I need help, Solve for Percent Error, Calculate the percent error if the experimental value of the density of go

ld is 18.7 g/cm^3 and the accepted value is 19.3 g/cm^3

1 answer:

Dima020 [189]3 years ago

7 0

Answer:

Step-by-step explanation:

so the steps to doing this are as follows :

Subtract the accepted value from the experimental value.

Take the absolute value of step 1.

Divide that answer by the accepted value.

Multiply that answer by 100 and add the % symbol to express the answer as a percentage.

|18.7-19.3|/19.3 = 0.031088 * 100% = 3.1%

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A total of 2n cards, of which 2 are aces, are to be randomly divided among two players, with each player receiving n cards. Each

klasskru [66]

Answer:

$P(X_s^c|X_F) =0.2$

$P(X_s^c|X_F) =0.31$

$P(X_s^c|X_F) =0.331$

Step-by-step explanation:

From the given information:

Let represent $X_F$ as the first player getting an ace

Let $X_S$ to be the second player getting an ace and

$\sim X_S$ as the second player not getting an ace.

So;

The probabiility of the second player not getting an ace and the first player getting an ace can be computed as;

$P(\sim X_S| X_F) = 1 - P(X_S|X_F)$

$P(X_S|X_F) = \dfrac{P(X_SX_F)}{P(X_F)}$

Let's determine the probability of getting an ace in the first player

i.e

$P(X_F) = 1 - P(X_F^c)$

$= 1 -\dfrac{(^{2n-2}_n)}{(^{2n}_n)}}$

$= 1 - \dfrac{n-1}{2(2n-1)}$

$= \dfrac{3n-1}{4n-2} --- (1)$

To determine the probability of the second player getting an ace and the first player getting an ace.

$P(X_sX_F) = \text{ (distribute aces to both ) and (select the left over n-1 cards from 2n-2 cards}$ $P(X_sX_F) = \dfrac{2(^{2n-2}C_{n-1})}{^{2n}C_n}$

$P(X_sX_F) = \dfrac{n}{2n -1}---(2)$

$P(X_s|X_F) = \dfrac{2}{1}$

$P(X_s|X_F) = \dfrac{2n}{3n -1}$

Thus, the conditional probability that the second player has no aces, provided that the first player declares affirmative is:

$P(X_s^c|X_F) = 1- \dfrac{2n}{3n -1}$

$P(X_s^c|X_F) = \dfrac{n-1}{3n -1}$

Therefore;

for n= 2

$P(X_s^c|X_F) = \dfrac{2-1}{3(2) -1}$

$P(X_s^c|X_F) = \dfrac{1}{6 -1}$

$P(X_s^c|X_F) = \dfrac{1}{5}$

$P(X_s^c|X_F) =0.2$

for n= 10

$P(X_s^c|X_F) = \dfrac{10-1}{3(10) -1}$

$P(X_s^c|X_F) = \dfrac{9}{30 -1}$

$P(X_s^c|X_F) = \dfrac{9}{29}$

$P(X_s^c|X_F) =0.31$

for n = 100

$P(X_s^c|X_F) = \dfrac{100-1}{3(100) -1}$

$P(X_s^c|X_F) = \dfrac{99}{300 -1}$

$P(X_s^c|X_F) = \dfrac{99}{299}$

$P(X_s^c|X_F) =0.331$

8 0

3 years ago

Maths help please!<br><br> Volume?

drek231 [11]

Answer:

I think it's 17 cm wish this helps.

4 0

3 years ago

3688÷6, how did I get this answer

Katena32 [7]

Answer:

1844/3

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A customer bought a car and paid $1,080 in sales tax.The rate is 6%?

Ghella [55]

The price of car is $18000

Step-by-step explanation:

Given

Sales tax on car = $1080

Rate = 6%

Let x be the price of the car

We know that as the rate of sale tax is 6%, 6% of x is 1080

Using mathematical notation

0.06x = 1080

Dividing both sides by 0.06

$\frac{0.06x}{0.06} = \frac{1080}{0.06}\\\\x = 18000$

Hence,

The price of car is $18000

Keywords: Rate of tax, Percentage

Learn more about percentae at:

#LearnwithBrainly

7 0

3 years ago

Two groups of students were asked how far they lived from their school. The table below shows the distances in miles

leonid [27]

<span>Group A (distance in miles) 1 1.5 3.03 3.2 2.8 1.5 1.8 2.5 2.2
Group B (distance in miles) 2 2.5 3.23 1.3 1.8 2.4 3 1.5 1.8

Group A mean = (1 + 1.5 + 3.03 + 3.2 + 2.8 + 1.5 + 1.8 + 2.5 + 2.2)/9 = 19.53/9 = 2.17
Group B mean = (2 + 2.5 + 3.23 + 1.3 + 1.8 + 2.4 + 3 + 1.5 + 1.8)/9 = 19.53/9 = 2.17

Therefore, the mean is equal for Group A and Group B.
</span>

7 0

3 years ago

Read 2 more answers