Answer:
a)
v=<400, 450+3t, 600-9.81t> m/s
r=<400t, 450t+3
/2, 15+600t-9.81
/2> m/s
b) attach file
c) t=122.324s
d) z=18369.9854 m
Step-by-step explanation:
a)
The equation for velocity of each component is v=
+at, in this case the x-axis component has acceleration 0
, the y-axis component has acceleration 3
and the z-axis component has -g
, where g=9.81
.
The equation for position of each component is x=
+
t+a
/2,
with <
,
,
>=<0,0,15>.
c) This time is t when z=0, with positive sign.
d)Solving the equation
=
+ 2*a*(z-
) for z, with
=0.
Let
be the random variable for the number of marks a given student receives on the exam.
10% of students obtain more than 75 marks, so
![P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10](https://tex.z-dn.net/?f=P%28X%3E75%29%3DP%5Cleft%28%5Cdfrac%7BX-%5Cmu%7D%5Csigma%3E%5Cdfrac%7B75-%5Cmu%7D%5Csigma%5Cright%29%3DP%28Z%3Ez_1%29%3D0.10)
where
follows a standard normal distribution. The critical value for an upper-tail probability of 10% is
![P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)](https://tex.z-dn.net/?f=P%28Z%3Ez_1%29%3D1-F_Z%28z_1%29%3D0.10%5Cimplies%20z_1%3DF_Z%5E%7B-1%7D%280.90%29)
where
denotes the CDF of
, and
denotes the inverse CDF. We have
![z_1=F_Z^{-1}(0.90)\approx1.2816](https://tex.z-dn.net/?f=z_1%3DF_Z%5E%7B-1%7D%280.90%29%5Capprox1.2816)
Similarly, because 20% of students obtain less than 40 marks, we have
![P(X](https://tex.z-dn.net/?f=P%28X%3C40%29%3DP%5Cleft%28%5Cdfrac%7BX-%5Cmu%7D%5Csigma%3C%5Cdfrac%7B40-%5Cmu%7D%5Csigma%5Cright%29%3DP%28Z%3Cz_2%29%3D0.20)
so that
![P(Z](https://tex.z-dn.net/?f=P%28Z%3Cz_2%29%3DF_Z%28z_2%29%3D0.20%5Cimplies%20z_2%3DF_Z%5E%7B-1%7D%280.20%29%5Capprox-0.8416)
Then
are such that
![\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma](https://tex.z-dn.net/?f=%5Cdfrac%7B75-%5Cmu%7D%5Csigma%5Capprox1.2816%5Cimplies75%5Capprox%5Cmu%2B1.2816%5Csigma)
![\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma](https://tex.z-dn.net/?f=%5Cdfrac%7B40-%5Cmu%7D%5Csigma%5Capprox-0.8416%5Cimplies40%5Capprox%5Cmu-0.8416%5Csigma)
and we find
![\mu\approx53.8739,\sigma\approx16.4848](https://tex.z-dn.net/?f=%5Cmu%5Capprox53.8739%2C%5Csigma%5Capprox16.4848)
Answer:
The fraction of one pound of cheese to be used in each sandwich is 1/8
Step-by-step explanation:
Here, we have a question that requires the method of dealing with the division of fractions.
To divide a fraction by a whole number, we multiply the denominator of the fraction by the whole number as follows;
![\frac{x}{y} \div z =\frac{x}{y\times z}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7By%7D%20%5Cdiv%20z%20%3D%5Cfrac%7Bx%7D%7By%5Ctimes%20z%7D)
Quantity of cheese Shane has = 1/2 pound
Number of sandwiches to be made = 4
Amount of cheese per sandwich = 1/2÷4 = 1/8.
Answer:
the one solution is -20
Step-by-step explanation:
-15x - 27 + 7 = -25 - 15x +5
-27 + 7 = -25 +5
-20 = -20
Answer:
1.3°F * 10 = 13°F
10.2°F
Step-by-step explanation:
Given that:
Temperature on the ground = - 2.8°F
Change in temperature per 1000 ft rise = 1.3°F
Change in temperature after the plane ascends 10000 ft
Change in temperature per 1000ft rise * (10000/1000)
1.3°F * 10 = 13°F
B.) Temperature at 10,000 feets
Temperature on ground + change in temperature at 10,000ft
-2.8°F + 13°F
= 10.2°F