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4 years ago

Someone help me please

Mathematics

1 answer:

ElenaW [278]4 years ago

4 0

i think this is the answer

Step-by-step explanation: So i think the answer is 2 : 3 and another one is

3 to 2

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A sample of 12 measurements has a mean of 16.5, and a sample of 15 measurements has a mean of 18.6. Find the mean of all 27 meas

Wewaii [24]

Answer: $17.67$

Step-by-step explanation:

Given

Sample of 12 measurements has a mean of 16.5 and

a sample of 15 measurements has a mean of 18.6

Take $\bar{x_1},n_1$ be the mean and no of measurements

and $\bar{x_2},n_2$ be the mean and no of measurements in second case

$\therefore \bar{x_1}=\dfrac{\sum a_1}{n_1}\\\\\Rightarrow \sum a_1=\bar{x_1}\times n_1\\\\\Rightarrow \sum a_1=198\quad \ldots(1)$

Similarly,

$\therefore \bar{x_2}=\dfrac{\sum a_2}{n_2}\\\\\Rightarrow \sum a_2=\bar{x_2}\times n_2\\\\\Rightarrow \sum a_2=279\quad \ldots(2)$

Mean of 27 measurements

$\Rightarrow \bar{x_3}=\dfrac{\sum a_1+\sum a_2}{n_1+n_2}\\\\\Rightarrow \bar{x_3}=\dfrac{198+279}{12+15}\\\\\Rightarrow \bar{x_3}=\dfrac{477}{27}\\\\\Rightarrow \bar{x_3}=17.67$

4 0

3 years ago

HELP MATH ASAP! I GIVE BRAINLIEST

aliya0001 [1]

The answer would be 0.

6 0

3 years ago

The reciprocal of the curvature is called

marissa [1.9K]

Answer:

the radius of curvature, R, is the reciprocal of the curvature.

8 0

3 years ago

If you save $125 each week for 12 weeks, how much will you be able to save?

Gre4nikov [31]

Answer:

1,500 dollars

Step-by-step explanation:

You would do 125 times 12 = 1,500

8 0

3 years ago

Read 2 more answers

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

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

We are looking for P(X<2)

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

In this case we want P(X>3)

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

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

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

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