### Complete the description of the piecewise function graphed below.

Complete the description of the piecewise function graphed below.

Solution:

summary:

f(x)=-2x-8 if -6<=x<=-3

f(x)=-3 if -3<x<=2

f(x)=(3x-14)/2 if 2<x<=6

### Discrete Example Suppose we have a 4-sided die and let X denote the random face that comes up on a throw. Its pmf is given by Table 1, where θ,p ∈ [0,1]. Suppose we throw

Question: Discrete Example Suppose we have a 4-sided die and let X denote the random face that comes up on a throw. Its pmf is given by Table 1, where θ,p ∈ [0,1]. Suppose we throw

the die a certain number of times and observe xi i’s, for i = 1,…,4 (i.e., face i comes up xi times).
(a) What is the likelihood of this experiment given θ? (You should treat p as a constant) (b) What is the maximum likelihood estimate of θ?

3.2 Normal Distribution

Suppose we sample n i.i.d. points from a Gaussian distribution with mean μ and variance σ?. Recall that the Gaussian pdf is given by

Compute the maximum likelihood estimate of parameters μ and σ?.

Solution:

### Suppose that a van with capacity for 7 passengers departs from a commuter station

Question: Suppose that a van with capacity for 7 passengers departs from a commuter station. Observation has shown that the bus never departs empty. Let X denote the number of passengers that are female and Y the number of passengers that are male. Assume that all possible (x, y) pairs are equally likely. (a) Find the probability that the number of female passengers is more than 5. (a) Find the probability that the total number of passengers is less than 5.

Solution:

pairs of passengers:-
{(0,7),(0,6),(0,5),(0,4),(0,3),(0,2),(0,1)
(1,6),(1,5),(1,4),(1,3),(1,2),(1,1),(1,0)
(2,5),(2,4),(2,3),(2,2),(2,1),(2,0)
(3,4),(3,3),(3,2),(3,1),(3,0)
(4,3),(4,2),(4,1),(4,0),
(5,2),(5,1),(5,0)
(6,1),(6,0)
(7,0)}
p(x,y) = 1/35 where x = no of female and y = no of males

a. Probability of female passengers is more than 5
possible outcomes = (6,0),(6,1),(7,0)
P(A) = 3/35 = 0.0857

b. Passenger is less than 5
possible outcomes = {(4,0),(3,0),(3,1),(2,2),(2,1),(2,0),
(1,3),(1,2),(1,1),(1,0),(0,3),(0,2),(0,1),(0,4)}
= 14
P(B) = 14/35 = 0.4

### Which statement below about asexual reproduction is FALSE?

Question: Which statement below about asexual reproduction is FALSE?
A. With asexual reproduction, offspring are genetically equivalent to the parent.
B. Asexual reproduction requires no partner.
C. Asexual reproduction requires meiosis.
D. Some organisms can reproduce both sexually and asexually.

Solution: Asexual reproduction requires meiosis.
only sexual reproduction required meiosis, as gamete fusion occurs, to maintain the stable chromosome number in the offspring due to this phenomenon, meiosis is required in sexual reproduction.

Question: If a diploid sperm fertilized a diploid egg, what would the result be?
A. Diploid (2n)
C. A mixture of diploid (2n) and quadruploid (4n)
D. Triploid (3n)

2n+2n = 4n.

Question: Which of the following represents things that are equivalent?
A. Two alleles for the same gene in a homologous chromosome pair
B. The sequences of DNA in the two sister chromatids of a chromosome before meiosis
C. The sequences of DNA in the two sister chromatids of a chromosome after meiosis
D. The number of homologous chromosomes per cell before and after meiosis

Solution: Two alleles for the same gene in a homologous chromosome pair.

Question: What problem would most likely occur if a haploid cell attempted to perform meiosis?
A. The cell could not replicate its DNA prior to meiosis.
B. The cell could not pair homologous chromosomes during meiosis I.
C. The cell would produce diploid daughter cells.

Solution: The cell could not pair homologous chromosomes during meiosis I.

Question: Which statement below is TRUE?
A. Crossing over shuffles genes between non-homologous chromosomes.
B. During melosis, cells arrange their alleles so that beneficial alleles are passed on together.
C. Without crossing over, offspring would be genetically identical to parents.
D. New combinations of alleles arise from the random events of crossing over, independent assortment, and mutation.

Solution: New combinations of alleles arise form the random events of crossing over, independent assortment, and mutation.

Question: What statement below is TRUE of the spindle during mitosis and meiosis?
A. The spindle always separates sister chromatids during anaphase.
B. The spindle always separates homologous chromosomes during anaphase.
C. Chromosomes are always attached to both spindle poles during metaphase.
D. The spindle always attaches to chromosomes at the kinetochore.

Solution: The spindle always attached to chromosomes at the kinetochore.

Question: If a person is heterozygous for the △32 allele of the CCR5 gene, how many of the four daughter cells produced by meiosis will have the △32 allele?
A. 1
B. 2
C. 4
D. Varies

Solution: 2.

Question: Which of the following processes has the effect of changing ploidy?
A. DNA replication
B. Meiosis I
C. Meiosis II
D. Mitosis

Solution: Meiosis I
During Anpahse I of meiosis I, as the chromosome segregation occurs, the ploidy will be reduced to half.

Question: The image below shows a homologous chromosome pair with two genes, A and B. Each of the two genes has two alleles (A1 and A2, B1 and B2). For this homologous pair of chromosomes, after completion of meiosis with NO CROSSING OVER, which of the following combinations of alleles might end up in one of the gametes?

A. A1 and B1
B. A2 and B1
C. A2 and B2
D. All of the above are possible.

Solution: A2 and B1

Question: For the same homologous pair of chromosomes shown in the previous question, which of the following combinations of alleles might end up in one of the gametes after completion of meiosis WITH CROSSING OVER?
A. A1 and B2
B. A2 and B1
C. A2 and B2
D. All of the above are possible

Solution: All of the above are possible.

### Signals and Systems Chapter 7, Problem 41AP

Question: In many practical situations a signal is recorded in the presence of an echo, which we would like to remove by appropriate processing. For example, in Figure P7.41 (a), we illustrate a system in which a receiver simultaneously receives a signal x(t) and an echo represented by an attenuated delayed replication of x(t). Thus, the receiver output is

Assume that x(t) is band limited [i.e., X(jω) = 0 for

and that | α | < 1.

(a) If

, and the sampling period is taken to be equal to determine the difference equation for the digital filter hire] such that yc(t) is proportional to x(t).

(b) With the assumptions of part (a), specify the gain A of the ideal lowpass filter such that

(c) Now suppose that

. Determine a choice for the sampling period T, the lowpass filter gain A, and the frequency response for the digital filter h[n] such that is proportional to x(t).

Solution:

(a) The received signal is,

It is ideally sampled with an impulse train of period,

.

Sampling with impulse functions is ideal sampling.

Where,

.

Translate the received signal to frequency domain.

The sampled signal in time domain is (or) the impulse train is,

,

It is a sequence of impulses with amplitudes equal to the sample value.

Remove the sampling period, as they simply correspond to a sample.

Convert the impulse train to a sequence.

Therefore, the sequence is,

The final output must be proportional only to

.

That implies the signal

must be a baseband signal and must be proportional to .

For the final output to be proportional to be only

; the frequency domain translation of the received signal is,

That means in the delayed version of

, needs to be eliminated, to get output exactly proportional to .

In frequency domain, this translates to elimination of the

term.

So the digital filter

must eliminate term.

The signal

is applied to the digital filter, .

The expression for output

is,

Convolution in time domain translates to multiplication in frequency domain.

Substitute for (since the output is proportional to input in time domain), for .

Thus, the frequency domain description of the digital filter is,

Determine the difference equation.

…… (1)

Apply inverse Fourier transform in equation (1).

Thus, the difference equation of the filter is

.

(b)

The output sequence

is converted to an impulse train.

That is, the sequence itself is an impulse train.

To recover the signal form this impulse train, the signal is passed through a low pass filter.

The output of the digital filter is taken to be exactly equal to .

Thus, the input to the low pass filter is

.

The low pass filter simply acts as a reconstruction filter.

The input is

, for the output of the ideal low pass filter to be exactly ,

Substitute

for .

Thus, the pass band gain of the ideal low pass filter is

.

c)

The impulse

Note that in this case,

.

The minimum sampling frequency, to avoid aliasing is,

.

The delay of the echo has a range of,

The sampling frequency of the sum of two signals is twice the bandwidth of the signal with the highest bandwidth.

The discrete sequence is,

…… (1)

The sampling frequency does not change if the signal is shifted.

The received signal is the sum of the signal

and the its shifted version .

The sampling frequency of sum of two signals is the twice the bandwidth of the signal with the highest bandwidth.

Thus, the sampling frequency is,

Thus, the sampling period must be ,

.

Apply Fourier transform to equation (1).

Determine the frequency response of the digital filter,

.

For the final output to be proportional to,

.

The only possibility is altering the function using

.

The input signal is,

Since the input is directly applied to the digital filter,

The frequency response of the output is multiplication of

and .

Substitute

for and for .

Thus, the gain of frequency response of the digital filter is,

The impulse train

is a sampled version of the signal ,due to the effect of digital filter.

For the final output to be exactly

, we need to pass it through a low pass filter.

The input to the ideal low pass filter is a sampled version of

.

So get the exact replica of the signal

, the gain of the ideal low pass filter must be .

Thus, the gain of the ideal low pass filter is

.

### Describe the physical changes that occur during frying hen eggs temperature 180°C

Question: 1. Describe the physical changes that occur during frying hen eggs temperature 180°C? at

2. Discuss how nitrites can be used to preserve cured meat products.

Solution:

Ans 1 Deep frying is an ancient process commonly used in food industry. It is a lost cost and highly demanding process which involves oil-food interaction at high temperature. When food is cook at high temperature dehydration occurs and food gets completely cooked because of which physical and chemical changes take place. There are vareity of examples such as starch gelatinization, protein denaturation, caramelization, maillard reaction ( giving food an aromatic and mouth watering aroma, which occurs when amino acids and sugar breaks down). Similarly deep frying of hen egg involves protein denaturation. The protein is a three dimensional arrangement of atoms in amino acid chain molecules and fold into spatial conformations via non colent interactions such as hydrogen bonding, ionic interactions and vanderwall forces. When the whole egg i.e yolk and white( albumen-major source of protein) is heated it turn into solid. This solidification of protein is known as coagulation. The heat causes the protein structure to lose water molecules and hence shrink them. Different proteins denature at diiferent temperature, similarly egg white starts denaturing at 61 degree celcius. As the protein changes its shape, it changes its taste and texture also.

Ans 2 Salts ( sodium chloride) are used as preservatives for meats from a long time.It was accidently found that nitrite salts increases the shelf life of meats by preventing rancidity( spoilage of food) and also the growth of bacteria. So now a days nitrite salts  such as sodium nitrites are being used as preservatives and purified nitrites are commercially manufactured for this purpose. Pure nitriles if consumed at levels of 3-5 gm can be dangerous and may result in death also as it binds with oxygen in body stronger than heamoglobin thus not allowing oxygen to reach the organs. So it is added to meat according to level(0.01-0.02% depending on type of meats) set by USDA. It helps in improving meat quality , when added at allowed levels set by USDA it inhibits the growth of almost all bacteria such as Clostridium botulinum, Clostridium perfringens, Listeria monocytogens.that caauses food spoilage and sickness.

### Please implement booth’s algorithm in logisim to solve signed multiplication.

Question:

Please implement booth’s algorithm in logisim to solve signed multiplication. The circuit should represent the one seen below.

The following inputs and outputs are required:

• Multiplicand: a 16-bit two’s complement input

• Multiplier: a 16-bit two’s complement input

• Mul (1-bit input): This input will be one if the instruction is the multiplication instruction

• Clock (1-bit input)

• Product: 32-bit two’s complement output

• M Ready (1-bit output): This output will be 1 if the product is ready

The inside circuit should use a register for the multiplier, the multiplicand, and the product.  A ROM circuit should control the clock signal and what happens on each clock edge. The ROM circuit triggers the registers. The ROM circuit can be seen below:

Solution:

Booth’s algorithm is a multiplication algorithm that multiplies two signed binary numbers in 2’s compliment notation.-

### Determine P(A1 I B1) and Determine P(A2 I B1)

Question:

A1     A2

B1 .4      .3

B2 .2      .1

Determine P(A1 I B1)

Determine P(A2 I B1)

Did your answers to parts (a) and (b) sum to 1? Is this a coincidence? Explain

Solution: a. P(A1 I B1) = P( A1 n B1)/ P(B1)

P( A1 n B1) = 0.4

P(B1) = 0.4+0.3 = .7

P(A1 I B1) = 0.4/0.7 = 0.5714

b. P(A2 I B1) = P( A1 n B1)/ P(B1)

P( A2 n B1) = 0.3

P(B1) = 0.4+0.3 = .7

P(A1 I B1) = 0.3/0.7 = 0.4285

### Tangent Lines Find the equation of the tangent line of the curve tan(2x+5y) = x at (0,0).

Question: Calculus: Tangent Lines Find the equation of the tangent line of the curve tan(2x+5y) = x at (0,0). Pick ONE option

y = 5x

y=x

y = x/5

y=-x/5

Solution:

We have to find the equation of the tangent line of the curve

tan(2x+5y)=x

at the point (0,0).

Now, to find this, we have to find the value of

which is the slope of the tangent line, at a given point.

Here,

Differentiating both sides with respect to x, we get

Putting the value of x=0 and y=0, we can say that

So, the slope of the tangent will be -(1/5).

Now, at the point (0,0), the equation of the straight line is

So, the correct answer is option (D) .

### The double riveted butt joint shown in the Figure connect two plates, which are each 2.5 mm thick

Question: The double riveted butt joint shown in the Figure connect two plates, which are each 2.5 mm thick, the rivet has a diameter of 3mm. If the failure strength of the rivets in shear is 370 N/mm2 and the ultimate tensile strength of the plate is 465 N/mm2 determine the necessary rivet pitch if the joint is to be designed so that failure due to shear in the rivets and failure due to tension in the plate occur simultaneously. Calculate the joint efficiency.

Solution:

Given that:
Thickness = 2.5 mm
Diameter = 3 mm
Tensile strength = 465 N/mm^2
Shear strength = 370 N/mm^2
Shearing of rivets can be calculated as below:
Ps = nπ/4 * d^2 * τ * 2 (double shear)
where
τ = shear strength
n = no of rivets pen pitch length = 2
d = 3
Now substitute in above

Ps = 2 π/4 * 3^2 * 370 * 2
Ps= 10461.5 N

Tearing resistance of plate:
Pt = (P-d)tθt
d = 3 mm
t = 2.5 mm
θt = 465 N/mm^2

Pt = (P – 3) * 2.5 * 465
As given question shearing resistance of rivet = tearing resistance of plate
that is
Ps = Pt
(P-3)2.5465 = 10462
(P-3) * 1162.5 = 10462
P – 3 = 10462/1162.5
P – 3 = 8.999
P = 8.99 + 3
P = 11.99
P = 12
Now the strength of the unriveted on solid plate pen pitch length
P = Ptθt
t = 2.5 mm
P = 12
θt = 465
Now substitute those value’s then
P = 122.5465
P = 13950N
where P= pitch of revert
Now to determine the efficiency of the joint we know that the
efficiency of joint = least of Ps,Pt/P
Now substitute those values
Ps = 10462N
P = 13950
= 10462/13950
= 0.749
Efficiency of joint = 74.9%