Answer:
The cosine of 86º is approximately 0.06976.
Step-by-step explanation:
The third degree Taylor polynomial for the cosine function centered at
is:

The value of 86º in radians is:


Then, the cosine of 86º is:


The cosine of 86º is approximately 0.06976.
<u>Answer-</u>
The standard error of the confidence interval is 0.63%
<u>Solution-</u>
Given,
n = 2373 (sample size)
x = 255 (number of people who bought)
The mean of the sample M will be,

Then the standard error SE will be,


Therefore, the standard error of the confidence interval is 0.63%
Answer:
Costo final= $412.38
Step-by-step explanation:
Dada la siguiente información:
Costo inicial= $355.5
Recargo de la tarjeta= 16% = 0.16
<u>Para calcular el costo final que debe pagar Silvia, debemos usar la siguiente información:</u>
Costo final= costo inicial*(1 + recargo)
Costo final= 355.5*1.16
Costo final= $412.38
First of all you have to find the missing measurements. The actual measurements for the angles in the hexagon are not given, but they give you an expression. You have to solve for x first so that you can plug it in and find the angle measurement. You have to equal the two sides that are given to you like this: 20x+48=33x+9. You solve for x and then plug it into each angle measurement. This should give you 108. Since it is a regular hexagon all of the sides are equal. If you look at the angle at the top of the hexagon you'll see two triangles and the angle. Since it lies on a straight line, it is all equal to 180. You already have the angle measurement of the hexagon and are missing the triangles. So 180-108=72. 72 is the missing part of the angle. You divide this by 2 in order to find each triangle angle measurements. the answer is 36 degrees.
Answer:
The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .
Step-by-step explanation:
Given as :
The mass of liquid water = 50 g
The initial temperature =
= 15°c
The final temperature =
= 100°c
The latent heat of vaporization of water = 2260.0 J/g
Let The amount of heat required to raise temperature = Q Joule
Now, From method
Heat = mass × latent heat × change in temperature
Or, Q = m × s × ΔT
or, Q = m × s × (
-
)
So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )
Or, Q = 50 g × 2260.0 J/g × 85°c
∴ Q = 9,605,000 joule
Or, Q = 9,605 × 10³ joule
Or, Q = 9605 kilo joule
Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer