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Andrei [34K]
3 years ago
15

Amanda has at most $40 to spend on flowers. She wants to buy a pair of red rose flowers for $18 dollars and spend the rest on li

ly flower. Each lily flower costs $11. Write an inequality for the number of lily flowers she can purchase
Mathematics
1 answer:
Mkey [24]3 years ago
5 0

Answer:

x ≤ 2

Step-by-step explanation:

2 red rose cost  $ 18

x flowers cost $ 11x

"At most "  means ≤.

11x + 18 ≤ 40

11x ≤ 40 - 18

11x ≤ 22

x ≤ 2

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What is the common multiples of 5,3,8
tangare [24]

Answer:

120.

Step-by-step explanation:

If you're asking for LCM the answer is 120.

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3 years ago
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What roles did militias play in the American Revolution? Your answer:
Annette [7]

Hey there! I'm happy to help!

A militia is a local army. During the Battle of Lexington and Concord, the local militia (called minutemen), the militia outnumbered the British at Concord and chased them all the way back to Boston. The militia aided in many American victories during the Revolutionary War.

I hope that this helps! Have a wonderful day! :D

8 0
3 years ago
(2 + 5i)(2 + i) <br><br> please help me to understand how to do it
Tpy6a [65]
<span>(2 + 5i)(2 + i) 
= 4 + 10i + 2i + 5i^2
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8 0
3 years ago
A pizza at papas cost $6.80 plus $0.90 for each topping.The cost of a pizza at little Italy’s is $7.30 plus $0.65 for each toppi
ira [324]

Answer:

At papa's it would be three topping  

at little Italy's it would be two topping

I hope this is correct sorry if I'm wrong

Step-by-step explanation:

5 0
3 years ago
the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai
ipn [44]

Answer:

The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.

Step-by-step explanation:

Use the trick of transforming the given random variable one that is standard normal distributed, aka a z-score, then look up the two percentile values in z-score tables to get two equations with two unknowns.

So, let x be the variable describing the marks of a student, and

z=(x-\mu)/\sigma

the standardized equivalent of that (mu - mean, sigma - standard deviation).

We are looking for values of mu and sigma. At this point we'd be out of luck, but, wait, we're given two bits of info: the 10% point (aka, 90th percentile) and the 20% point (can be interpreted as 100-20th percentile). For each point we can use z-score tables to look up the corresponding values of z (just search for z tables). I found:

z-score for the 10% point: z_10=1.28

z-score for the 20% point: z_20=-0.84

That gives us two equations:

z_{10}=1.28=(75-\mu)/\sigma\\z_{20}=-0.84=(40-\mu)\sigma

and can be solved for mu and sigma (do the work on your end, I am showing my result):

\mu=53.87\,\,,\,\, \sigma=16.51

The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.

3 0
3 years ago
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