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Gennadij [26K]
4 years ago
15

Assuming the point estimate in Problem 6.36 is the true population parameter, what is the probability that a particular assay, w

hen expressed in the log scale to the base 2, is no more than 1.5 log units off from its true mean value for a particular woman?

Engineering
1 answer:
sp2606 [1]4 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability is 0.958

Explanation:

The explanation is shown on the second and third uploaded image

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An insulated tank having a total volume of 0.6 m3 is divided into two compartments. Initially one compartment contains 0.4 m3 of
IRISSAK [1]

Answer:

a. The final temperature is 69.8°C

b. The final pressure is 2.67bar

c. The amount of entropy produced is 0.38568kJ/K

Explanation:

Pls, check the attached files for detail step by step explanation as typing the solution here can not be explicit.

5 0
3 years ago
Solve the inequality. Then graph your solution.<br> -9v – 10 &lt; 7y +6
Cerrena [4.2K]
16

if you add 9+10 you get 18 - 7+6
5 0
3 years ago
Is a compass a analog or a digital sensor?
ollegr [7]
A compass is a analog sensor
5 0
3 years ago
one number is 11 more than another number. find the two number if three times the larger number exceeds four times the smaller n
vaieri [72.5K]

Answer:

a = 40

b = 29

Explanation:

Give a place holder for the numbers that we don't know.

Lets call the two numbers a and b.

From the given info, we can write an expression and solve it:

"one number is 11 more than another number"

a = 11 + b

from this, we know that a > b.

''three times the larger number exceeds four times the smaller number by 4"

3a = 4b + 4

Now we have 2 equations, we can use them to solve using whatever method you want.

a = 11 + b

3a = 4b + 4

I will be using matrices RREF to solve for this.

a - b = 11

3a - 4b = 4

\begin{bmatrix}1 & -1  & 11\\3 & -4 & 4 \end{bmatrix}

\begin{bmatrix}1 & 0  & 40\\0 & 1 & 29 \end{bmatrix}

a = 40

b = 29

6 0
3 years ago
An electrical current of 700 A flows through a stainlesssteel cable having a diameter of 5 mm and an electricalresistance of 610
KatRina [158]

Answer:

778.4°C

Explanation:

I = 700

R = 6x10⁻⁴

we first calculate the rate of heat that is being transferred by the current

q = I²R

q = 700²(6x10⁻⁴)

= 490000x0.0006

= 294 W/M

we calculate the surface temperature

Ts = T∞ + \frac{q}{h\pi Di}

Ts = 30+\frac{294}{25*\frac{22}{7}*\frac{5}{1000}  }

Ts=30+\frac{294}{0.3928} \\

Ts =30+748.4\\Ts = 778.4

The surface temperature is therefore 778.4°C if the cable is bare

6 0
3 years ago
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