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3 years ago

The results of inspection of DNA samples taken over the past 10 days are given below. Sample size is 100.

Mathematics

1 answer:

ASHA 777 [7]3 years ago

7 0

Answer:

The average defective rate = .057

Step-by-step explanation:

Given -

Sample size n = 100

Day - 1 2 3 4 5 6 7 8 9 10

Defectives - 7 6 6 9 5 6 0 8 9 1

The average defective rate is the no of defective in each day for past ten days

The average defective rate = $\frac{\sum no\; of \; effective \; in\; each\; day}{no\; of \; day \times n}$

= $\frac{7 + 6 + 6 + 9 + 5 + 6 + 0 + 8 + 9 + 1}{10\times 100}$

= $\frac{57}{1000}$

= .057

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1/4 + 1/2 + x = - 3/4 what is x equal to

ludmilkaskok [199]

Answer:

-6/4 or -1 1/2

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3 years ago

Estimate the sum of 487 and 310

Jobisdone [24]

500+300=800 hope this helped:)

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3 years ago

Read 2 more answers

For the differential equations dydx=sqrt(y^2−36) does the existence/uniqueness theorem guarantee that there is a solution to thi

julsineya [31]

Answer:

1. (-4,6) there is no a solution to the equation through this point

2. (2,−6) there is no a solution to the equation through this point

3. (−5,39) there is a solution to the equation through this point

4. (−1,45) there is a solution to the equation through this point

Step-by-step explanation:

Using the existence and uniqueness theorem:

Let:

$F(x,y)=\sqrt{y^2-36} \\\\and\\\\\frac{\partial F}{\partial y} =\frac{y}{\sqrt{y^2-36} }$

Now, let's find the domain of $F(x,y)$ , due to the square root:

$y^2-36 \geq 0\\\\y^2\geq 36\\\\y \geq 6\hspace{12}or\hspace{12}y\leq-6$

So the domain of the function is:

$y \in R\hspace{12}y\geq6\hspace{12}or\hspace{12}y\leq-6$

Now, due to the fraction $\frac{\partial F}{\partial y}$ the denominator must be also different from 0, so:

$y^2-36\neq0\\\\y \neq \pm6$

So, the theorem tells us that for each $y_0\in R:\hspace{12}y_0>6\hspace{12}or\hspace{12}y_0$ there exists a unique solution defined in an open interval around $x_0$ .

1. (-4,6) there is no a solution to the equation through this point because $y_0=6$

2. (2,−6) there is no a solution to the equation through this point because

$y_0=-6$

3. (−5,39) there is a solution to the equation through this point because

$y_0>6$

4. (−1,45) there is a solution to the equation through this point because

$y_0>6$

6 0

3 years ago

Which function , A or B has a greater rate of change ? Be sure to include the values for the rates of change in your answer . Ex

arsen [322]

<h2>Answer:</h2>

A rate of change tells us how one quantity changes in relation to another quantity. In a mathematical language, we can write this as follows:

$Rate \ of \ Change=\frac{Change \ in \ y}{Change \ in \ x}$

For linear functions. the rate of change is the slope of the line. Thus:

FOR THE TABLE:

By taking two points we can get the rate of change, so let's take $(1,5) \ and \ (2,7)$ :

$ROC=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ ROC=\frac{7-5}{2-1} \\ \\ ROC=\frac{2}{1} \\ \\ \boxed{ROC=2}$

FOR THE GRAPH:

Let's take $(1,1) \ and \ (2,4)$ :

$ROC=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ ROC=\frac{4-1}{2-1} \\ \\ ROC=\frac{3}{1} \\ \\ \boxed{ROC=3}$

As you can see, $3>2$ so <em>the ROC of the function given by the graph is greater than the ROC of the function given by the table.</em>

3 0

3 years ago

suppose that a severe virus in impacting a bee colony and the population of bees has a half life of 5 days What percentage of th

BartSMP [9]

So we want to know what percentage of bees would survive a virus after 19 days of decay if the bee population has a half life of 5 days. So every 5 days 50% of bees die. Lets say that N is the number of bees before the virus. So after 5 days is 50% less bees, so 50% of N remains. After another 5 days again, 50% bees die, and 50% out of 50% is 25%. So after 10 days, 25% of bees remain or 25%N or 0.25*N. After another 5 days its 12.5 % of bees remain or 0.125*N. And after 4 days 40% more bees die. And that is 0.4*0.125*N = 0.05N. 0.05*100% is 5%. So after 19 days, 5% of bees remain and 95% of bees is dead.

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3 years ago