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polet [3.4K]
3 years ago
5

How do you divide 56$ in a ratio of 5:3? What would be the answer?

Mathematics
1 answer:
Sati [7]3 years ago
8 0
Sum the ratio:

s = 5 + 3 = 8

to get the first amount:

a1 = (3/8) * 56 = 21 $

a2 = (5/8) * 56 = 35 $

Check:

total = 21 + 35 = 56 $ ( as given)

ratio = 35/21 = 5/3 (as required)


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Monique bought a shirt for $22.80 during a 30% off sale. How much does the shirt cost when it is not on sale?
spin [16.1K]

Answer:

29.64

Step-by-step explanation:

First we multiply 0.3 with 22.8.This will give us 6.84 which is the 30% off.Now we add 22.80 and 6.84 and we will get our answer 29.64.

4 0
2 years ago
Read 2 more answers
5/6 of 15 i need help????????????
emmainna [20.7K]

Answer: 12.5 or 12 1/2

Step-by-step explanation:

7 0
2 years ago
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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
What is the sum of the first 37 terms of the arithmetic sequence?
lidiya [134]

Answer:

The sum of the first 37 terms of the arithmetic sequence is 2997.

Step-by-step explanation:

Arithmetic sequence concepts:

The general rule of an arithmetic sequence is the following:

a_{n+1} = a_{n} + d

In which d is the common diference between each term.

We can expand the general equation to find the nth term from the first, by the following equation:

a_{n} = a_{1} + (n-1)*d

The sum of the first n terms of an arithmetic sequence is given by:

S_{n} = \frac{n(a_{1} + a_{n})}{2}

In this question:

a_{1} = -27, d = -21 - (-27) = -15 - (-21) = ... = 6

We want the sum of the first 37 terms, so we have to find a_{37}

a_{n} = a_{1} + (n-1)*d

a_{37} = a_{1} + (36)*d

a_{37} = -27 + 36*6

a_{37} = 189

Then

S_{37} = \frac{37(-27 + 189)}{2} = 2997

The sum of the first 37 terms of the arithmetic sequence is 2997.

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=5x%20%5Ctimes%2021%20%3D%207" id="TexFormula1" title="5x \times 21 = 7" alt="5x \times 21 = 7"
pentagon [3]
Answer: Should be X= 1/15
4 0
2 years ago
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