Find the angle measure to the nearest degree. tan V = 7.1154

How much heat in kJ is released when nitrogen and hydrogen react as below to form 100.g ammonia heat equals-92.2kj

tresset_1 [31]

100g of ammonia releases 271.18 kJ of heat.

<u>Explanation:</u>

Given:

N₂ reacts with H₂

The reaction is:

2N₂ + 3H₂ → 2NH₃ + 92.2 kJ

According to the balanced reaction:

2 X 17g of ammonia releases 92.2 kJ of heat

34g of ammonia releases 92.2 kJ of heat

100g of ammonia will release = $\frac{92.2}{34} X 100$ kJ of heat

= 271.18 kJ

Therefore, 100g of ammonia releases 271.18 kJ of heat.

8 0

3 years ago

(-5z)^3 using only positove exponents

Elena L [17]

Answer:

Shown below

Step-by-step explanation:

First allocate the power of three to every number

$-5^{3} * z^{3}$ = $-125 * z^{3}$

not sure what you mean by positive exponents, the exponents already are positive. Maybe you mean positive exponents and integers?

8 0

2 years ago

Sample annual salaries (in thousands of dollars) for employees at a company are listed. 42 36 48 51 39 39 42 36 48 33 39 42 45 (

Sonja [21]

Answer:

a) Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45

$\bar X = 41.538$

$s = 5.317$

b) 44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25

$\bar X = 43.615$

$s= 5.583$

c) 3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75

$\bar X= 3.462$

$s = 0.443$

d) As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3.462).

And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.

And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.

Step-by-step explanation:

For this case we can use the following formulas for the mean and standard deviation:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Part a

Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45

And if we calculate the mean we got:

$\bar X = 41.538$

$s = 5.317$

Part b

For this case we know that each value present a 5% of rise so we just need to multiply each value bu 1.05 and we have this new dataset:

44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25

And if we calculate the new mean and deviation we got:

$\bar X = 43.615$

$s= 5.583$

Part c

The new dataset would be each value divided by 12 so we have:

3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75

And the new mean and deviation would be:

$\bar X= 3.462$

$s = 0.443$

Part d

As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3,462).

And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.

And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.

5 0

4 years ago

Is the following expression true or false? [x^2 + 8x + 16] · [x^2 – 8x + 16] = (x2 – 16)^2

Vilka [71]

$(x^{2} +8x+16)(x^{2} -8x+16)\\(x+4)^{2} (x-4)^{2} \\(x^{2}-16)^{2} \\$