If it takes a robot 2/5 hours to do 3 tasks how long does it take to do 1 task

For the following pairs of assertions, indicate which do not comply with our rules for setting up hypotheses and why (the subscr

torisob [31]

Answer:

Options B, C, D & E do not comply with the rules for setting up hypotheses

Step-by-step explanation:

(A). H0: m = 100, Ha: m > 100.

These hypotheses fully complies with our rules.

B). H0: s = 20, Ha: s 20.

H0 is not an equality claim such as σ = 20. Hence, these hypotheses are not in compliance with our rules.

(C). H0: p -:: .25, Ha: p = .25

H0 should definitely contain the equality claim, although Ha contains it here. Thus, these are not legitimate and don't comply with our rules

(D)H0: m1 - m2 = 25, Ha: m1 - m2 > 100

The given value of 25 for m1 - m2 in H0 should also appear in Ha,but it does not do so here, so it doesn't comply with our rules.

E)H0: S2 = S2, H : S2 -:: S2

Each S2 is a statistic and so it does not belong in a hypothesis.

8 0

4 years ago

ILL GIVE BRAINLIEST

anygoal [31]

Answer:

First option: f(x) is greater

5 0

3 years ago

Read 2 more answers

The Royal Fruit Company produces two types of fruit drinks. The first type is 55% pure fruit juice, and the second type is 80% p

Stolb23 [73]

Answer:

First type of fruit drinks: 48 pints

Second type of fruit drinks: 32 pints

Step-by-step explanation:

Let's call A the amount of first type of fruit drinks. 55% pure fruit juice

Let's call B the amount of second type of fruit drinks. 80% pure fruit juice

The resulting mixture should have 65% pure fruit juice and 80 pints.

Then we know that the total amount of mixture will be:

$A + B = 80$

Then the total amount of pure fruit juice in the mixture will be:

$0.55A + 0.8B = 0.65 * 80$

$0.55A + 0.8B = 52$

Then we have two equations and two unknowns so we solve the system of equations. Multiply the first equation by -0.8 and add it to the second equation:

$-0.8A -0.8B = -0.8*80$

$-0.8A -0.8B = -64$

+

$0.55A + 0.8B = 52$

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

$-0.25A = -12$

$A = \frac{-12}{-0.25}$

$A = 48\ pints$

We substitute the value of A into one of the two equations and solve for B.

$48 + B = 80$

$B = 32\ pints$

6 0

3 years ago

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

Which expressions are equilvalent to 3^x

SVETLANKA909090 [29]

Where’s the question?

8 0

3 years ago