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3 years ago

During a flu epidemic, 44 students out of the 80 who are in Ms Grant’s class were absent. What percent were absent?

Mathematics

2 answers:

wel3 years ago

8 0

44/80 x 100
=0.55 x 100
=55%

maw [93]3 years ago

5 0

The answer would be 54%

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Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

3 years ago

Protease is an enzyme that breaks down proteins. What is the end product of this protein breakdown? A. amino acids B. fatty acid

mezya [45]

Answer:

A protease (also called a peptidase or ) is an enzyme an enzyme that catalyzes (increases the rate of) proteolysis, the breakdown of proteins into smaller polypeptides or single amino acids. They do this by cleaving the peptide bonproteins by hydrolysis, a reaction where water breaks bonds.

6 0

3 years ago

How to write this numbers in word 30.659<br>

Karo-lina-s [1.5K]

Answer:

thirty and six hundred fifty-nine thousandths

Step-by-step explanation:

6 0

3 years ago

A man is 1.8m tall and standing 60m away from a light pole and the sun cast his shadow at 6m. What is the height of the light po

dsp73

<span>19.8 m We have 2 similar triangles. For the 1st triangle, we have two legs, one being the man's height of 1.8m and the second being the length of his shadow at 6m. So we have the ratio 1.8/6. The other triangle is comprised of the pole's height and length of the pole's shadow which will be 60m + 6m = 66meters. So we have another ratio of x/66 which has the same value as the first ratio. So let's solve for x. 1.8/6 = x/66 66*1.8/6 = x 11*1.8 = x 19.8 = x So the light pole is 19.8 meters tall.</span>

6 0

3 years ago

A man cut his 2 fingers and add 4 hands with 7 fingers. And he also chop his leg 3 times everyday till January to December. How

Olenka [21]

Answer: 30

Step-by-step explanation:

10 is the started fingers of the hands

2 is fingers, 4 are the hands, 7 the fingers of the hands

2 + (4 · 7) = 30

-3 is the legs per day, and 365 are the days per year

-3 · 365 = -1095

this is a bizarre question

4 0

3 years ago