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OleMash [197]
4 years ago
11

Can someone please help?

Mathematics
2 answers:
aleksklad [387]4 years ago
8 0

Answer:

yes i can

Step-by-step explanation:

EleoNora [17]4 years ago
8 0

Answer:

sure why not mark brainliest

Step-by-step explanation:

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The output of a chemical process is continually monitored to ensure that the concentration remains within acceptable limits. Whe
marissa [1.9K]

Answer:

a)  0.31 = 31%

b) 0.03 = 3%

c) 0.36 = 36%

d) 2 times

Step-by-step explanation:

If F_X(x) is the cumulative distribution function of the random variable X, then by definition the probability P of the random variable is given by

P(X \leq x) = F_X(x)

If additionally the random variable is discrete (only has non-negative integers as outcomes as is the case in this problem) then

P(X=a)=F_X(a)-\lim_{x \to a^-}F_X(x)

a)

We are looking for P(X<2)

P(X < 2) = P(X\leq 2)-P(X=2)=F_X(2)-\lim_{x \to 2^-}F_X(x)=0.84-0.53=0.31

b)

In this case we want P(X>3)

P(X >3) = 1-P(X\leq 3)=1-F_X(3)=1-0.97=0.03

c)

Now, we are interested in P(X=1)

P(X =1) =F_X(1)-\lim_{x \to 1^-}F_X(x)=0.53-0.17=0.36

d)

The expected number of times that the process is recalibrated during the week is the expected value of the probability distribution:

P(X=1)+2P(X=2)+...+nP(X=n)+...

But it is easy to see that P(X=n) = 0 if n is an integer >4

So, the expected value is

P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)

We already have P(X=1) and P(X=2). Let's compute the rest

P(X =3) =F_X(3)-\lim_{x \to 3^-}F_X(x)=0.97-0.84=0.13

P(X =4) =F_X(4)-\lim_{x \to 4^-}F_X(x)=1-0.97=0.03

and the expected value is

0.36 + 2*0.53+3*0.13+4*0.03= 1.93 = 2 times rounding to the nearest integer.

8 0
4 years ago
The mean is:
nata0808 [166]

Answer:

A

Step-by-step explanation:

Because the mean has to describe

4 0
3 years ago
A cube has a volume of 216 cubic feet. What is the length of an edge of this cube?
kupik [55]

Answer:

6

Step-by-step explanation:

a cube has sides of the same length hence its volume is obtain as cubed side.

we have the volume and we need the side:

\sqrt[3]{216} =\sqrt[3]{3^{3} 2^{3} } = 3*2 = 6

3 0
3 years ago
five silver coins weigh 125gm and are worth $6. Ten bronze coins weigh 500gm and are worth 80 cents. A number of silver and bron
Kisachek [45]

Answer:

100 silver coins and 175 bronze


Step-by-step explanation:

let x be the number of silver coins and y the number of bronze coins.  

Worth of these coins together = 6x/(5)+0.08/(10)y

=1.2x+0.008y

Weight of one silver coin = 125/5 = 25 gm

and one bronze coin = 500/10 = 50 gms.

Weight of these coins together = 25x + 50y

Equate these two items i.e. value and weight to actuals given.

i.e. 1.2x+0.08y = 134 and    ... i

25x+50y = 11 kg 250 gm = 11250 gms. ... ii

Divide the second equation by 25

x+2y = 450   ...  iii

Multiply I equation by 25

30x+2y =3350   ... iv

iii-iv gives -29x = -2900 or

x =100

Substitute in x+2y = 450

2y = 350 or y =175

6 0
3 years ago
Help pls for brainliest
KIM [24]

Answer:

8x + 2x (4)

Step-by-step explanation:

If you are trying to simplify or isolate the variables, its 4. Please tell me if this answer is wrong

8 0
3 years ago
Read 2 more answers
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