Answer:
<em>The goal is to persuade and inform and to provide personality so I would say </em><em>alliteration </em><em>and</em> <em>amplification</em><em>, but that depends on the writer. Alliteration is the use of tongue-twisters to create a positive mood. Amplification repeats the same words to grab the reader's attention.</em>
The second continental congress raised an army
All you have to do is to find the following from the z-score table:
On the 1st row of the table, you can read the value of z & you complete the value by reading the corresponding horizontal line. However watch out which table you are using & what is requested from you. For example in the following question:
a. Between the mean and a z = 1.27: , if you use a normal table it will give you the value of ==> 0.8990==>89.8%, which is the area from the EXTREME LEFT TO z=1.27., whereas yo have to calculate the area from the mean to
to z=1.27 0r 0.8990 -.5=0.398 or 39.85 (.5, because you start counting from the mean):
so:
a. Between the mean and a z = 1.27 ==> 39.8%
b. Between the mean and a z = -0.91 ==> 0.31859==>18.31%
c. Between the z scores of -1.84 and 1.39 . In this case you will find the value of each z & you will add up both: for z= -1.84 ==> 0.88482 or 88.48%
d. Between the z scores of -2.58 and 2.58: same as above,==>99.1%
e. Above a z of 1.96 , means from z to the end. The area from the mean to z=1.96 ==> 0.475, what wee need is from this point to the end
that is 0.5-0.475=2.5%
f. Below a z score of -2.58 same as above but -2.58 to the extreme left
for z=-2.58 ==>0.49506 , & below this area==> 0.5-0.49506==> 0.49%
g. Above a z score of -1.72 ==> 95.72%
h. Below a z of 1.96 ==>97.5%
To solve your problem, we can apply Newton's second law of motion.
Fⁿ = ma
<span>Fnet = net force acting on the skier </span>
<span>m = mass of the skier = 53 kg </span>
<span>a = acceleration up the slope </span>
<span>Since velocity is constant then acceleration is zero, hence the above equation becomes </span>
<span>Fnet = 0 </span>
<span>Fnet = F - Wx - f </span>
<span>where </span>
<span>F = force exerted by the tow bar </span>
<span>Wx = component of skier's weight parallel to the incline = 53(9.8)(sin 20) </span>
<span>f = frictional force = 0.160(53)(9.8)(cos 20) </span>
<span>Substituting appropriate values, </span>
<span>F - 53(9.8)(sin 20) - 0.160(53)(9.8)(cos 20) = 0 </span>
<span>and solving for F, </span>
<span>F = 99.55 N </span>
<span>Hope this helps and have a nice day!</span>
