What is 3.14 times 10

Answer:

a) Reject H0 if tcalc > 1.7960

b)    

c-1) ii. FALSE

c-2) ii.FALSE

c-3)i. TRUE

Step-by-step explanation:

1) Data given and notation  

represent the mean time for the sample  

represent the sample standard deviation for the sample  

sample size  

represent the value that we want to test

represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

represent the p value for the test (variable of interest)  

a) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 5 days, the system of hypothesis would be:  

Null hypothesis:  

Alternative hypothesis:  

We don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

 (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Rejection zone

On this case we need a critical value that accumulates 0.05 of the area on the right tail. The degrees of freedom are given by 11. And we can use the following excel code to find the critical value : "T.INV(1-0.95,11)" and the critical value would be given by .

And the rejection zone is given by:

Reject H0 if tcalc > 1.7960

b) Calculate the statistic

We can replace in formula (1) the info given like this:  

P-value

The first step is calculate the degrees of freedom, on this case:

Since is a one side test the p value would be:  

c-1. The null hypothesis should be rejected.

ii. FALSE

c-2. The average repair time is longer than 5 days.

ii. FALSE

Conclusion  

If we compare the p value and the significance level given we see that so we can conclude that we have enough evidence to fail reject the null hypothesis, and the true mean is not significantly higher than 5.  

c-3 At α = .05 is the goal being met?

i. TRUE

We fail to reject the null hypothesis so then the goal is met.

Answer: first blank: 1

second blank: 9

third blank: -7

fourth blank: 2

1x-9=-7

x=2

Answer:

Mean= $21.5067

Median = $15.8

Standard deviation= $19.02

Coefficient of skewness= $0.8991

Step-by-step explanation:

Mean =( $4.0 +$6.0 +$7.4+ $10.6 +$12.5+ $13.6+ $15.4+ $15.8 +$16.8

+$17.4+ $19.1 +$22.3+ $37.1 +$43.2 +$81.4)/15

Mean =$ 322.6/15

Mean= $21.5067

Median= middle number

Median = $15.8

Variance=( ($4.0-.$21.5)²+( $6.0. -.$21.5)²+( $7.4 -.$21.5)²+( $10.6 -.$21.5)²+( $12.5 -.$21.5)²+( $13.6. -.$21.5)²+ ($15.4 -.$21.5)²+( $15.8 -.$21.5)²+ ( $16.8 -.$21.5)²+ ($17.4-.$21.5)² +($19.1 -.$21.5)²+ ($22.3 -.$21.5)²+ ($37.1 -.$21.5)²+ ($43.2-.$21.5)²+( $81.4-.$21.5)²)/15

Variance=$ 5424.79/15

Variance=$ 361.65

Standard deviation= √ variance

Standard deviation= √361.65

Standard deviation= $19.02

Coefficient of skewness

=3( mean-median)/standard deviation

= 3(21.5-15.8)/19.02

= 3(5.7)/19.02

= 17.1/19.02

Coefficient of skewness= 0.8991

Answer:

Probability = 0.119

Step-by-step explanation:

P (Coronavirus) = P(Person online & corona) or P(Person offline & corona)

(0.70 x 0.02) + (0.30 x 0.35)

0.014 + 0.105

0.119

Answer:

Altitude of the pile will increase by 16.56 cm per second.

Step-by-step explanation:

Sand is poured onto a surface at the rate = 13 cm³ per second

Or

It forms a conical pile with a diameter d cm and height of the pile = h cm

Here d = h

Volume of the pile cm³per sec.

Since h = d = 2r [r is the radius of the circular base]

r =

Since cm³per sec.

13 = [For h = 1 cm]

cm per second.

Therefore, altitude of the pile will increase by 16.56 cm per second.

Final answer:

To solve this problem, we first find the expression for the volume of the cone in terms of the height. We then differentiate this expression to get the relation between the rates of change of the volume and the height. By substituting the given values, we can find the rate of change of the height when the cone is 1 cm high.

Explanation:

The question is related to the application of calculus in Physics, specifically rates of change in the context of real-world problem involving a three dimensional geometric shape - a cone. The student asks how fast the altitude of a pile of sand is increasing at a given time if sand is being poured onto a surface at a constant rate and the pile forms a cone whose base diameter is always equal to its altitude.

We know that the volume of a cone is given by V = (1/3)πr²h, where r is the radius of the base and h is the altitude. Since in this problem the base diameter is always equal to its altitude, we have d = 2r = h, or r = h/2.

Replace r in the volume formula, yielding V = (1/3)π(h/2)²h = (1/12)πh³. Differentiate this expression with respect to time (t) to find the rate of change of V with respect to t, dV/dt = (1/4)πh² * dh/dt.

Given that sand is poured at a constant rate of 13 cm³/sec (that is, dV/dt = 13), we can solve for dh/dt when h = 1cm. Substituting the given values into the equation, 13 = (1/4)π(1)² * dh/dt, we find dh/dt = 13/(π/4) = 52/π cm/sec. Therefore, when the conical pile is 1 cm high, the altitude is increasing at a rate of 52/π cm/sec.

Learn more about Calculus in Physics here:

brainly.com/question/28688032

#SPJ3

Answer:

3

Step-by-step explanation:no