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MakcuM [25]
3 years ago
13

A manufacturer makes three types of screws. Type A comes in a bulk pack of 1000 screws. Type B comes in a bulk pack of 500 screw

s. Type C comes in a bulk pack of 800 screws. Type A screws each have a probability of .001 of having a flaw. The probability is .003 for Type B and .005 for Type C.
(a) If a contractor buys one pack of each type of screws, find the expected total number of defective screws among the three packs.
(b) Find the probability that there are no more than 5 defective screws total among the three packs.
(c) Find the expected number of packs that have no more than 5 defective screws. (Your answer will be between 0 and 3.)
Mathematics
1 answer:
Tpy6a [65]3 years ago
8 0

Answer:

Step-by-step explanation:

Given that a manufacturer makes three types of screws

Type             A        B           C Total

   

Pack         1000 500   800  

P for flaw 0.001 0.03 0.005  

Pack*p                 1 15                    4     20

a) Expected total number of defective screws = 20

b) For one pack each no of defective screws =20

Hence for 5 expected number number of packs form each should be 1/3

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sergey [27]

Answer:

$793.80

Step-by-step explanation:

You want to calculate the interest on $630 at 6.5% interest per year after 4 year(s).

The formula we'll use for this is the simple interest formula, or:

<em>I = P x r x t</em>

<em>Where: </em>

<em> </em>

<em>P is the principal amount, $630.00. </em>

<em>r is the interest rate, 6.5% per year, or in decimal form, 6.5/100=0.065. </em>

<em>t is the time involved, 4....year(s) time periods. </em>

<em>So, t is 4....year time periods. </em>

<em>To find the simple interest, we multiply 630 × 0.065 × 4 to get that: </em>

<em> </em>

<em>The interest is: $163.80 </em>

<em>Usually now, the interest is added onto the principal to figure some new amount after 4 year(s), </em>

<em>or 630.00 + 163.80 = 793.80. For example: </em>

<em> </em>

<em>If you borrowed the $630.00, you would now owe $793.80 </em>

<em>If you loaned someone $630.00, you would now be due $793.80 </em>

<em>If owned something, like a $630.00 bond, it would be worth $793.80 now.</em>

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Test yourself 2
mafiozo [28]
Ok, so dy/dx=0 at the point (0,3) that is where x=0 and y=3.

\int { 6x+6dx } \\ \\ =\frac { 6{ x }^{ 2 } }{ 2 } +6x+C\\ \\ =3{ x }^{ 2 }+6x+C

\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =3{ x }^{ 2 }+6x+C

Now, f'(x)=0 when x=0.

Therefore:

0=C\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =3{ x }^{ 2 }+6x

Now:

\int { 3{ x }^{ 2 } } +6xdx\\ \\ =\frac { 3{ x }^{ 3 } }{ 3 } +\frac { 6x^{ 2 } }{ 2 } +C

={ x }^{ 3 }+3{ x }^{ 2 }+C\\ \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+C

But when x=0, y=3, therefore:

3=C\\ \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+3
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saw5 [17]

From the graph, we get that:

The domain is (-4,4).

The function has no zeros.

The function is positive in the interval (-4, 0].

The function is negative in the interval (0,4).

------------------------------------

The domain of a function is the set that contains all the possible input values. In a graph, it is the values of the x-axis, that is, the horizontal axis.

In this question, the values of x are in the interval of (-4,4) thus, the domain is (-4,4).

The zeros are the values of x for which y = 0, that is, the values of x where the function crosses the horizontal axis. In this question, the function does not cross the horizontal axis, thus, it has no zeros.

The function is positive when the graph is above the horizontal axis. In this question, it is in the interval (-4,0].

The function is negative when the graph is below the horizontal axis. In this question, it is in the interval (0,4).

A similar problem is given at brainly.com/question/24493729

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