3 years ago

What is the value of |–3.25| ?

Mathematics

1 answer:

34kurt3 years ago

3 0

Answer:

3.25

Step-by-step explanation:

Rule: An absolute value will always be positive.

=> Referring to the rule, the absolute value of |–3.25| is 3.25.

Hoped this helped.

$BrainiacUser1357$

You might be interested in

Which list puts the following things in correct order from smallest to largest?

slega [8]

Answer:

A. "hope this helps"

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the value of y so that r⎹⎹ s A. 32 B. 56 C. 64 D. 124 please help

4vir4ik [10]

We need more detail I will help when i get a real question

7 0

3 years ago

A company uses three different assembly lines- A1, A2, and A3- to manufacture a particular component. Of thosemanufactured by li

hammer [34]

Answer:

The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

Step-by-step explanation:

The three different assembly lines are: A₁, A₂ and A₃.

Denote R as the event that a component needs rework.

It is given that:

$P (R|A_{1})=0.05\\P (R|A_{2})=0.08\\P (R|A_{3})=0.10\\P (A_{1})=0.50\\P (A_{2})=0.30\\P (A_{3})=0.20$

Compute the probability that a randomly selected component needs rework as follows:

$P(R)=P(R|A_{1})P(A_{1})+P(R|A_{2})P(A_{2})+P(R|A_{3})P(A_{3})\\=(0.05\times0.50)+(0.08\times0.30)+(0.10\times0.20)\\=0.069$

Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:

$P (A_{1}|R)=\frac{P(R|A_{1})P(A_{1})}{P(R)}=\frac{0.05\times0.50}{0.069} =0.3623$

Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

6 0

3 years ago

Ignore this

marissa [1.9K]

Answer:

Thanks for free

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the surface area of the regular dodecahedron. PLEASE HELP!!

kakasveta [241]

12*123.9=1486.8
plz rate brainlyist

5 0

3 years ago