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lukranit [14]
2 years ago
12

A water bottle that holds 24 1/2 ounces of water is 3/5 full. Is there enough water in the bottle to fill three glasses with 5 o

unces each?
Mathematics
2 answers:
Lelu [443]2 years ago
5 0

Answer:

No, there is not enough water.

Step-by-step explanation:

The water bottle holds 24.5  and there is 3/5 of it.

3/5 of 24.5 is 14.7 ounces

You want to fill three glasses with 5 ounces each. Multiply 3 x 5 which gives you 15.

14.7 is less than 15

Subtract 15 from 14.7 and you'll have a negative or not enough so no, it cannot fill three glasses with 5 each.

nekit [7.7K]2 years ago
4 0
No there is not enough water
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D) y= 3x^2 + 2 represents a function.

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A train traveling at a constant speed takes a passenger from Deer Lake to Liberty, stopping first at Summerville and next at Cyp
nasty-shy [4]
Deer Lake - Summerville - Cypress - Liberty
Deer Lake to Summerville is 3
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3 years ago
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A local clothing store sells jeans and shirts. The store makes a $10 profit on the sale of jeans and $3 profit on the sale of th
Zepler [3.9K]

Answer:

An inequality to model the situation is :   ( 10 m +  3 n)  ≥  150

Here, m : The number of jeans sold

          n : The number of shirts sold.

Step-by-step explanation:

Here, the profit on each sale of jeans = $ 10

The profit on each sale of shirt = $ 3

The profit needed to be made = AT LEAST $150

Now, here: Let us assume the number of jeans sold  = m

And, The number of shorts sold  =n

So, the profit on m jeans = m  x ( Profit on 1 jeans) = m x ( $ 10)  = 10 m

So, the profit on n shirts = n  x ( Profit on 1 shirt) = n x ( $ 3)  = 3 n

⇒ The total profit on m jeans + n shirts  = ( 10 m +  3 n)

But, according to the question, the profit should AT LEAST $150.

⇒ TOTAL PROFIT  ≥ $ 150

⇒  ( 10 m +  3 n)  ≥  150

Hence, an inequality to model the situation is :   ( 10 m +  3 n)  ≥  150

3 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.

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3 years ago
Which words in the sentence are the complete noun clause?
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D
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