# For the below questions, consider a consumer that consumes two goods, x and z with the following utility function. u with bar on top space equals space x to the power of 1 third end exponent z to the power of 2 over 3 end exponent suppose initial values for income and the prices of goods x and z are y equals 90, p subscript x equals space 10, and p subscript z equals 15 respectively, then the price of good x falls to syntax error from line 1 column 89 to line 1 column 100. unexpected '\'.. what is the magnitude of the total effect

(a) 25 V

(b) 15 V

(c) 0 V

(d) -25 V

Explanation:

Notice that the electric field is completely in the x direction (i hat in it expression) so there is only one component of the electric field which is Ex = 500 V/m. We take V0 (at x = 0, y = 0 and z = 0) to be our reference potential.

Recall that in a uniform field the potential (V) is defined as the product of the electric field (E) times the distance (d):

V = E * d

then we use the different values of d (converting them to meters first) to estimate the potential difference at different locations:

(a) at x = +5.00 cm, y = 0, z = 0:

V = Ex * 0.05 = 500 * 0.05 = 25 V

(b) at x = +3.00 cm, y = +4.00 cm, z = 0:

V = Ex * 0.03 = 500 * 0.03 = 15 V (Notice that Ey and Ez are zero, so they don't contribute)

(c) at x = 0, y = +5.00 cm, z = 0 (notice that for the x component of the field, the displacement is 0, giving V = 0 V, and also the y and z components of the field are zero, so they also render zero for the potential.

(d) at x = -5.00 cm, y = 0, z = 0; (again Ey and Ez don't bring any contribution), and V = Ex * (-0.05) = - 25 V

a) 59.98

b) 2.99

c) 2.99

d) Significantly High

Step-by-step explanation:

Part a)

Highest speed measured = x = 75.6 Mbps

Average/Mean speed = = 15.62 Mbps

Standard Deviation = s = 20.03 Mbps

We need to find the difference between carrier's highest data speed and the mean of all 50 data speeds i.e. x -

x - = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between carrier's highest data speed and the mean of all 50 data speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

Using the given values, we get:

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the carrier's highest data speed is significantly higher.

EF = BC = a ÐF is a right angle. FD = CA = b triangle EF = BC = a angle F is a right angle. FD = CA = b In triangle DEF, By Pythagoras Theorem, a2 + b2 = c2 the given AB=c= a^2 + b^2 square root Theorefore AB = DE But by construction, BC = EF and CA = FD triangle ABC congruent to DEF (S.S.S

We have given that Δ ABC is similar to Δ CBD and Δ ABC is similar to Δ ACD according to the attached picture.

Because of the similarity the corresponding sides are proportional. Then a/c = f/a

and b/e = c/b. If we cross multiply, we'll get

a² = cf and b² = ce

If we add these equalities together, we'll get

a² + b² = c (f + e)

From the beginning, we know that c = e+f

Then, the final result is:

a² + b² = c²

lets say we have a triangle abc and it's right triangle because it has 90 degrees angle. the longest side of the angle or the opposite side of a right angle ab is a hypotenuse.

lets call the angle ac as segment a

the angle bc as segment b

the angle ab (hypotenuse) as segment c

now i want to show the between them by constructing another segment between c and the hypotenuse, i want them to intersect at the right angle and call another new point as a d.

the reason why i did that because now i can show between similar triangles.

now we have three triangle

triangle abc

triangle dbc

triangle adc

adc is the similar to abc because both of them have a right angle and both share angle dac

△adc~△acb

ac/ab = ad/ac (and if we rewrite it properly)

a/c = d/a (cross multiply)

a² = cd

△bdc~△bca

bc/ba = bd/bc (and if we rewrite it properly)

b/c=e/b (cross multiply)

b² = ce

now a² + b² = cd + ce

cd + ce = c (d+e)

* if d and e is a part of a segment c our equation is actually

c(c) = c²

now our is a² +b² =c² we just proved the pythagorean theorem.