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stich3 [128]
3 years ago
8

A quality-control inspector is testing a batch of printed circuit boards to see wheater they are capable of performing in a high

temperature environment. He knows that the boards that will survive will pass all five of the tests with probability 98%. They will pass at least four tests with probability 99%, and they always pass at least three. On the other hand, the boards that will not survive sometimes pass the tests as well. In fact, 2% pass all five tests, and another 20% pass exactly four. The rest pass at most three tests. The inspector decides that if a board passes all five tests, he will classify it as "good." Otherwise, he'll classify it as "bad." The manager says that the probability of a type I error must be no larger than 0.01.
a. What does a type II error mean here?
b. What is the probability of a type II error?
Mathematics
1 answer:
avanturin [10]3 years ago
6 0

Answer:

see explaination

Step-by-step explanation:

Here the null hypothesis is that the PCB survives against the alternate that the PCB 'does not survive'. The test says that the PCB will survice if it is classified as 'good'; or, it will not survive if it is classifies as 'bad'.

a. The Type II error is the error committed when a PCB which cannot actually survive is classified as 'good'.

b. Therefore P(Type II error) = P(The PCB is classified as 'good' | PCB does not survives) = 0.03.

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Please fix the question and help me do it the right way
Anestetic [448]

the last step is wrong because we multiply the powers in this case we do not subtract them.

we use power subtract in division.

we use power multiplication when we raise to power and the root is power like square root of x

\sqrt{x}  =  {x}^{ \frac{1}{2} }

so

\sqrt[3]{x}  =  {x}^{ \frac{1}{3} }

in our case

this step is wrong

2x \times  \sqrt[3]{ \frac{2}{y}  \times  \frac{y}{y} }

it should be

2 x \times  \sqrt[3]{ \frac{2}{y}  \times  \frac{ {y}^{2} }{ {y}^{2} } }

so we get

2x \times  \sqrt[3]{2 {y}^{2} \times  \frac{1}{ {y}^{3} }  }

=  \frac{2x}{y}  \times  \sqrt[3]{2 {y}^{2} }

3 0
2 years ago
A bag contains 120 marbles. Some are red and the rest are black.There are 19 red marbles for every black marble.How many red mar
Charra [1.4K]

Answer:

there are 114

Step-by-step explanation:

Ratio of red marbles to black marbles is 19:1. (Data from 3rd sentence)

19+1= 20

20 units= 120 (Data from 1st sentence)

1 unit= 120/20= 6

19 units= 120-6= 114 (Values solved in working above)

Ans: There are 114 red marbles.

6 0
2 years ago
Read 2 more answers
If m(x)=x+5/x-1 and n(x)=x-3 which function has the same domain as (m•n)(x)
worty [1.4K]
Consider the functions m and n:


i) m(x)=\frac{x+5}{x-1}   for x≠1.

the domain of m is  all the real numbers except 1.


ii) n(x)=x-3 is a polynomial function, because x-3 is a linear polynomial. 

The domain of any polynomial function is all real numbers. To determine the Range of this function, let c be a value in the range:

x-3=c, then x=3+c.

so if we want the function to produce c, we just set x=3+c, check:

n(3+c)=(3+c)-3=c.

This means that the Range of n is all real numbers.


What this means, is that the input (x) of the function n(x) can be any real number, and the output (n(x)) can also be any number.

so let n(x)=a, consider m(a):

m(a)=(a+5)/(a-1), clearly a≠1, this means n(x)=x-3≠1, that is x≠-4
 

This means that the domain of the composed function, m(n(x)) =  say f(x), 

is all Real numbers except 4.


Check: letting x be 4, would mean n(4)=4-3=1

which would mean m(n(4))=m(1)=(1+5)/(1-1)=6/0, which makes no sense.


Answer: Any function with domain R-{4}, has the same domain of (mon)(x) 

6 0
3 years ago
Is the point (3, 4) a solution to the equation 3y = 2x +6 ? Explain why using evidence from your work
irinina [24]

Answer:

Yes

Step-by-step explanation:

When graphing it you can see that (3, 4) is clearly a solution. The proper equation using slope intercept form is y= 2/3x + 2.

7 0
3 years ago
A person invests $4000 at 2% interest compounded annually for 4 years and then invests the balance (the $4000 plus the interest
faltersainse [42]
\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
A=4000\left(1+\frac{0.02}{1}\right)^{1\cdot 4}\implies A=4000(1.02)^4\implies A\approx 4329.73

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396

add both amounts, and that's her investment for the 11 years.
7 0
3 years ago
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