Can you please help me thank yo

Salaries of entry-level computer engineers have Normal distribution with unknown mean and variance. Three randomly selected comp

avanturin [10]

Answer:

H0 : μ = 60000

H1 : μ ≠ 60000

Test statistic = 3.464

Step-by-step explanation:

Given :

Sample mean salary, xbar = 80000

Sample standard deviation, s = 10000

Population mean salary , μ = 60000

Sample size, n = 3

Hypothesis :

H0 : μ = 60000

H1 : μ ≠ 60000

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (80000 - 60000) ÷ (10000/√(3))

T = 20000 / 5773.5026

T = 3.464

The Decison region :

If Tstatistic >Tcritical

Tcritical at 10%, df = 2 ; two - tailed = 2.9199

Tstatistic > Tcritical ; He

5 0

3 years ago

Evaluate 7 sigma n=1 2(-2)^n-1

oksian1 [2.3K]

\[\sum_{n=1}^{7} 2(-2)^{n-1}\]

5 0

3 years ago

Read 2 more answers

La suma de tres números pares consecutivos es 60. hallar sus números

telo118 [61]

19, 20, & 21 (add them up)

6 0

4 years ago

Read 2 more answers

Consider a machine that fills 275 gallon tanks of water. The owner discovers that the machine does not always dispense exactly 2

fenix001 [56]

Answer:

(a) The random variable X is a continuous random variable.

(b) The probability density function is shown below.

(c) The probability that a 275 gallon tank gets less than 272.5 gallons or more than 273.8 gallons of water is 0.285.

Step-by-step explanation:

Let the random variable X be denoted as the amount of water filled by the machine in a 275 gallon tank.

(a)

The random variable X is a continuous random variable.

A continuous random variables assumes infinite values. That is, they assume values in a fixed interval. For example, the distance covered by a car.

A discrete random variable assumes fixed definite values. They assume whole number values. For example, number of customers visiting a bank in an hour.

The amount of water in the tank can be any value between 0 to 275 gallon.

Hence, the random variable X is a continuous random variable.

(b)

The probability density function of the continuous random variable X is given as follows:

$0.25;\ 272$

$f_{X}(x) =0.20;\ 273$

$0.55;\ 274$

(c)

Compute the value of P (272.5 < X < 273.8) as follows:

$P(272.5$

Thus, the probability that a 275 gallon tank gets less than 272.5 gallons or more than 273.8 gallons of water is 0.285.

4 0

3 years ago

Need help and explain please!!

lukranit [14]

Answer:

$x=-4\text{ and } x=3$

Step-by-step explanation:

We are given the second derivative:

$g''(x)=(x-3)^2(x+4)(x-6)$

And we want to find its inflection points.

To do so, we will first determine possible inflection points. These occur whenever g''(x) = 0 or is undefined.

Next, we will test values for the intervals. Inflection points occur if and only if the sign changes before and after the point.

So first, finding the zeros, we see that:

$0=(x-3)^2(x+4)(x-6)\Rightarrow x=-4, 3, 6$

So, we can draw the following number-line:

<----(-4)--------------(3)----(6)---->

Now, we will test values for the intervals x < -4, -4 < x < 3, 3 < x < 6, and x > 6.

Testing for x < -4, we can use -5. So:

$g^\prime^\prime(-5)=(-5-3)^2(-5+4)(-5-6)=704>0$

Since we acquired a positive result, g(x) is concave up for x < -4.

For -4 < x < 3, we can use 0. So:

$g^\prime^\prime(0)=(0-3)^2(0+4)(0-6)=-216$

Since we acquired a negative result, g(x) is concave down for -4 < x < 3.

And since the sign changed before and after the possible inflection point at x = -4, x = -4 is indeed an inflection point.

For 3 < x < 6, we can use 4. So:

$g^\prime^\prime(4)=(4-3)^2(4+4)(4-6)=-16$

Since we acquired a negative result, g(x) is concave down for 3 < x < 6.

Since the sign didn't change before and after the possible inflection point at x = 3 (it stayed negative both times), x = -3 is not a inflection point.

And finally, for x > 6, we can use 7. So:

$g^\prime^\prime(7)=(7-3)^2(7+4)(7-6)=176>0$

So, g(x) is concave up for x > 6.

And since we changed signs before and after the inflection point at x = 6, x = 6 is indeed an inflection point.

3 0

3 years ago