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Lilit [14]
3 years ago
10

EMB agar is a medium used in the identification and isolation of pathogenic bacteria. It contains digested meat proteins as a so

urce of organic nutrients. Two indicator dyes, eosin and methylene blue, inhibit the growth of gram-positive bacteria and distinguish between lactose fermenting and nonlactose fermenting organisms. Lactose fermenters form metallic green or deep purple colonies, whereas the nonlactose fermenters form completely colorless colonies. EMB agar is an example of which of the following?A selective medium, a differential medium, and a complex medium
Engineering
1 answer:
Ilya [14]3 years ago
5 0

Answer:

A selective medium, a differential medium, and a complex medium.

Explanation:

A selective media is a microbiological media which only support the growth of a particular specie or types of species of microorganisms,this media acts in such a way to inhibit or hinder the growth of other microorganisms.

Differential media are media that acts to Identifying particular strains of microorganisms of similar species.

Complex media are media used for the growth of microorganisms this which contains complex or a wide range of nutrients with chemical composition which may be difficult to determine.

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Explain the benefits of reading A vocabulary section at the beginning of a text
Zinaida [17]

Answer: Deatiled information

Explanation:

It helps the reader organize and sort information in the text. It provides detailed information about the content of the text. It prepares readers to look for new vocabulary as they read.

3 0
4 years ago
Read 2 more answers
On the first statistics exam, the coefficient of determination between the hours studied and the grade earned was 80%. The stand
True [87]

Answer:

\left[\begin{array}{ccccc}&DF&SS&MS&F\\Regression&1&7200&7200&72\\Error&18&1800&100\\total&19&900\end{array}\right]

Explanation:

Sample size, n=20

Degrees of freedom is 1

Number of degrees of freedom for error is n-2 hence 20-2=18

Total number of degrees of freedom is 18+1=19

Standard error estimate is s_{y-x}=\sqrt {\frac {SSE}{n-2}}

Here, SSE=(n-2)s_{y-x}^{2}=(20-2)(10)^{2}=1800

Coefficient of determination r^{2}=\frac {SSE}{SS total}

Here, SSR=r^{2}(SSR+SSE)

SSR=\frac {r^{2}}{1-r^{2}} SSE=\frac {0.8}{1-0.8}(1800)=7200

The total sum of squares is

SS total=SSR+SSE=7200+1800=9000

MSR=SSR=7200

MSE=\frac {SSE}{n-2}=\frac {1800}{20-2}=100

F value is given by

F=\frac {MSR}{MSE}=\frac {7200}{100}=100

The ANOVA table is then  

\left[\begin{array}{ccccc}&DF&SS&MS&F\\Regression&1&7200&7200&72\\Error&18&1800&100\\total&19&900\end{array}\right]

4 0
4 years ago
Consider a car manufacturing firm, with more than 100 facilities worldwide. What would be the positive aspects of having high nu
liberstina [14]
More work getting done and faster
7 0
3 years ago
Read 2 more answers
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
Iteru [2.4K]

Answer:

% increase = 26.32%

Explanation:

From conservation of mass, we can say that;

Mass flow rate at inlet = mass flow rate at exit.

Thus;

m'1 = m'2

Formula for mass flow rate is;

m' = ρV'

Where V' is volumetric flow rate = Av

Thus;

m' = ρAv

Where;

ρ is density

A is area

v is velocity

Therefore from m'1 = m'2, we can say that;

ρ1•A1•v1 = ρ2•A2•v2

Since the duct has a constant diameter, then A1 = A2

Thus, we now have;

ρ1•v1 = ρ2•v2

Making v2 the subject, we have;

v2 = ρ1•v1/ρ2

Now, since we want to find the percent increase in the velocity of the air as it flows through the dryer,we would use;

% increase = ((v2 - v1)/v1) × 100%

We have v2 = ρ1•v1/ρ2

Thus;

% increase = ((ρ1•v1/ρ2) - v1)/v1) × 100%

Factorizing v1 out, we have;

% increase = ((ρ1/ρ2) - 1)/1) × 100%

We are given;

ρ1 = 1.2 kg/m³

ρ2 = 0.95 kg/m³

Thus;

% increase = ((1.2/0.95) - 1)/1) × 100%

% increase = 26.32%

8 0
3 years ago
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E 5 73 GPa and an ultimate strength of 140
NNADVOKAT [17]

Answer:

81.76 N/mm² ( MPa), 1.71233

Explanation:

Modulus of elasticity = stress / strain

stress = modulus of elastic × strain

strain = ΔL / L = 250.28 mm - 250 mm / 250 mm = 0.00112

Modulus of elasticity E = 73 GPa = 73 × 10³ MPa where 1 MPa = 1 N/mm²

E = 73 × 10³N/mm²

stress =  73 × 10³N/mm²× 0.00112 = 81.76 N/mm² ( MPa)

b) Factor of safety = maximum allowable stress / induced stress = 140 MPa / 81.76 MPa = 1.71233

8 0
4 years ago
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