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Verizon [17]
2 years ago
14

If with 150 grams of ground meat a tortilla is prepared to make a hamburger. How many tortillas can I make with one and a half k

ilograms of ground meat?​
Mathematics
2 answers:
Rashid [163]2 years ago
4 0

Answer:

You can make 10 tortillas with one and a half kilograms of ground meat

Step-by-step explanation:

one and a half kilograms to grams

1.5kg x 1000 = 1500g

150 grams prepared to make 1 hamburger

than, 150 x h = 1500

h = 1500 / 150

h = 10

Bond [772]2 years ago
4 0

Answer:

10 tortillas

Step-by-step explanation:

First convert kilograms to grams:

   1 kg = 1000 g

⇒ 1.5 kg = 1500 g

Let x = number of tortillas made with 1500 g of meat

Set up a ratio of tortillas and meat and solve for x:

⇒ 1 tortilla : x tortillas = 150 g meat : 1500 g meat

\sf \implies \dfrac{1}{x}=\dfrac{150}{1500}

\sf \implies 1 \cdot 1500= 150 \cdot x

\sf \implies 1500=150x

\implies \sf x=\dfrac{1500}{150}

\implies \sf x=10

Therefore, <u>10 tortillas</u> can be made with 1.5 kg of ground meat.

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3 years ago
The scores on the LSAT are approximately normal with mean of 150.7 and standard deviation of 10.2. (Source: www.lsat.org.) Queen
faltersainse [42]

Answer:

a=150.7 -0.385*10.2=146.773

So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.  

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

Let X the random variable that represent the  scores on the LSAT of a population, and for this case we know the distribution for X is given by:

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Where \mu=150.7 and \sigma=10.2

We want to find a value a, such that we satisfy this condition:

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P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.35 of the area on the left and 0.65 of the area on the right it's z=-0.385. On this case P(Z<-0.385)=0.35 and P(Z>-0.385)=0.65

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

Z=-0.385

And if we solve for a we got

a=150.7 -0.385*10.2=146.773

So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.  

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