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4 years ago

7

Write a program that prompts the user to enter two characters and display the corresponding major and year status. The first cha

racter indicates the major. And the second character is a number character 1, 2, 3, 4, which indicates whether a student is a freshman, sophomore, junior, or senior. We consider only the following majors:
B (or b): Biology

C (or c): Computer Science

I (or i): Information Technology and Systems

Note that your program needs to let the user know if the major or year is invalid. Also, your program should be case-insensitive: your program should tell "Biology" either user type ‘b’ or ‘B’.

1 answer:

lara31 [8.8K]4 years ago

7 0

Answer:

Following is attached the images for source code and the result according to given question.

I hope it will help you a lot!

Explanation:

You might be interested in

Based on the pattern, what are the next two terms of the sequence? 9,94,916,964,9256,... A. 91024,94096 B. 9260,91028 C. 9260,92

Hatshy [7]

Answer:

The answer is "Option A".

Explanation:

Series:

$9, 94, 916, 964, 9256, ........$

Solving the above series:

$\to 9\\ \to 9(4) =94\\\to 9 (4^2) = 9(16) =916\\\to 9 (4^3) = 9(64) =964\\\to 9 (4^4) = 9(256) =9256\\\to 9 (4^5) = 9(1024) =91024\\\to 9 (4^6) = 9(4096) =94096\\$

So, the series is: $9, 94, 916, 964, 9256, 91024, 94096, .................$

6 0

3 years ago

Write a program that asks the user to input a vector of integers of arbitrary length. Then, using a for-end loop the program exa

ELEN [110]

Answer:

%Program prompts user to input vector

v = input('Enter the input vector: ');

%Program shows the value that user entered

fprintf('The input vector:\n ')

disp(v)

%Loop for checking all array elements

for i = 1 : length(v)

%check if the element is a positive number

if v(i) > 0

%double the element

v(i) = v(i) * 2;

%else the element is negative number.

else

%triple the element

v(i) = v(i) * 3;

end

end

%display the modified vector

fprintf('The modified vector:\n ')

disp(v)

4 0

4 years ago

B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget

kap26 [50]

Answer:

minimum current level required = 8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = $\frac{\pi }{4} * D^2 * l$ = $\frac{\pi }{4} * 5.5^2 * 5$ = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 ) = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED use the relation below

electrical energy = I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore 8975.95 = I ( current )

4 0

3 years ago

An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc

Vadim26 [7]

Answer: 78.89%

Explanation:

Given : Sample size : n= 1200

Sample mean : $\overline{x}=2.45$

Standard deviation : $\sigma=0.07$

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, $z=\dfrac{x-\mu}{\sigma}$ , the z-value corresponds to 2.39 will be :-

$z=\dfrac{2.39-2.45}{0.07}\approx-0.86$

z-value corresponds to 2.60 will be :-

$z=\dfrac{2.60-2.45}{0.07}\approx2.14$

Using the standard normal table for z, we have

P-value = $P(-0.86$

$=P(z$

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

7 0

3 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago