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2 years ago

Was this a fair test? is it representative of the effectiveness of the test substance

1 answer:

8 0

No. The test is not a fair test. The test is artificial. Testing the disinfectant against bacteria in a test tube is not representative of the effectiveness of the disinfectant against microbes in the environment. Test tube tests test for efficacy. Effectiveness can only be tested out in the environment.

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PLEASE HELP!! WILL GIVE BRAINLIEST!!!

Murrr4er [49]

Answer:

option D

Step-by-step explanation:

Jerry solved this equation:

$3(x-\frac{1}{4})=\frac{13}{6}$

In the first step , she multiplied 3 inside the parenthesis

1. $3x-\frac{3}{4}=\frac{13}{6}$

In step 2, 3/4 is added on both sides

2. $3x-\frac{3}{4}+\frac{3}{4}=\frac{13}{6}+\frac{3}{4}$

In step 3, we take LCD 12

3. $3x=\frac{26}{12}+\frac{9}{12}$

In step 4, add the fractions

4. $3x=\frac{35}{12}$

Here, 3 or 3/1 are same.

$\frac{3}{1}x=\frac{35}{12}$

In step 5, to remove 3/1 we multiply both sides by 1/3.

Instead of multiplying 1/3 , Jerry made an error by multiplying 3/1

$\frac{1}{3} \cdot \frac{3}{1}x=\frac{1}{3} \cdot \frac{35}{12}$

$x=\frac{35}{36}$

In step 5, he should have multiplied both sides by 1/3

7 0

2 years ago

Read 2 more answers

Let e f (x) = x2 – 2x - 4 What is the average rate of change for the quadratic function from X = -1 tox to x = 42 Enter your ans

sammy [17]

Answer:

The average rate of change of the given function

A(x) = 1

Step-by-step explanation:

Step(i):-

Given function f(x) = x² - 2x -4

And given that x = a = -1 and x=b = 4

The average rate of change of the given function

$A(x) = \frac{f(b)-f(a)}{b-a}$

Step(ii):-

f(x) = x² - 2x -4

f(-1) = (-1)² - 2(-1) -4 = 1+2-4 = -1

f(4) = 4² -2(4) -4 = 16 -12 = 4

The average rate of change of the given function

$A(x) = \frac{f(b)-f(a)}{b-a}$

$A(x) = \frac{4-(-1)}{4-(-1)} = \frac{5}{5} = 1$

final answer:-

The average rate of change of the given function

A(x) = 1

5 0

2 years ago

Read 2 more answers

How am I supposed to integrate this?

svlad2 [7]

$6x^2-4x+1=6\left(x-\dfrac13\right)^2+\dfrac13\ge0$

which means the parabola lies above the x-axis over its entire domain. This means the area is given by

$\displaystyle\int_{-1}^2(6x^2-4x+1)\,\mathrm dx=2x^3-x^2+x\bigg|_{x=-1}^{x=2}=10-(-5)=15$