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BARSIC [14]
3 years ago
12

Instructions: Find the missing side lengths. Leave your answers as radicals in simplest form.

Mathematics
1 answer:
gogolik [260]3 years ago
3 0

Answer:

a = 4

b = 4

Step-by-step explanation:

This is a special right triangle with angle measures of 45° 45° 90° and side lengths x x x√2

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The adjustment of the thickness of the lens to make close vision possible is ca​
Savatey [412]

Answer: Accommodation

Step-by-step explanation:

Accommodation is the adjustment of the optics of the eye to keep an object in focus on the retina as its distance from the eye varies.

4 0
3 years ago
a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade
son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})

P(at least 1 dull blade)=1-0.2931

P(at least 1 dull blade)=0.7068

5 0
3 years ago
#3Elizabeth takes her younger siblings to the fair. She spends a total of
Debora [2.8K]
Take 40.00 divide by 8 to pay 5.00 a bag
5 0
3 years ago
The surface area of this prism is _______ cm squared
Katen [24]
Can you help me with my posted questions
6 0
3 years ago
The total cost to rent 5 chairs and 3 tables is $27. The total cost to rent 2 chairs and 12 tables is $81. What is the cost to r
kati45 [8]

The cost to rent each chair is $1.5 and cost to rent each table is $6.5

<h3>Applications of systems of linear equations </h3>

From the question, we are to determine the cost to rent each chair and each table

Let c represent chair

and

t represent table

From the given information,

The total cost to rent 5 chairs and 3 tables is $27

That is,

5c + 3t = 27 ------------ (1)

Also,

The total cost to rent 2 chairs and 12 tables is $81

That is,

2c + 12t = 81 ---------- (2)

Now, solve the equations simultaneously

5c + 3t = 27 ------------ (1)

2c + 12t = 81 ---------- (2)

Multiply equation (1) by 2 and multiply equation (2) by 5

2 × [5c + 3t = 27 ]

5 × [2c + 12t = 81 ]

10c + 6t = 54        ------------- (3)

10c + 60t = 405   ------------- (4)

Subtract equation (4) from equation (3)

10c + 6t = 54        

10c + 60t = 405

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

-54t = -351

t = -351/-54

t = 6.5

Substitute the value of t into equation (2)
2c + 12t = 81

2c + 12(6.5) = 81

2c + 78 = 81

2c = 81 - 78

2c = 3

c = 3/2

c = 1.5

∴ The cost of chair is $1.5 and cost of table is $6.5

Hence, the cost to rent each chair is $1.5 and cost to rent each table is $6.5

Learn more on Solving system of linear equations here: brainly.com/question/13729904

#SPJ1

8 0
2 years ago
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