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zhuklara [117]
3 years ago
15

Mrs. Perry is taking a group of students to the Clark Planetarium where tickets cost $7 each. The bus will cost $50 and she has

a budget of $200. How many people can attend under these constraints?​
Mathematics
1 answer:
kenny6666 [7]3 years ago
7 0
Answer:
71 people can attend under these constraints

Explanation:

So you write the slope intercept form equation which is 7x+50=200
You got 7x because every tickets cost 7
You add 50 because of the bus
And you’re wondering how many people are able to go by just spending about $200

So you subtract 50 on both side and you are left with 7x=150 so you have to divide 7 on both side next and you get about 71.4... so basically just round it up to 71 people
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Jonas weight 63 pounds. jane weight 67 pounds. round each number to the nearest ten.
IrinaVladis [17]

Answer:

Jonas = 60 pounds

Jane = 70 pounds

Remember- 5 or more, raise the score. 4 or less, let it rest.

Jonas is 63 pounds. 3 is less than 4, so rounded to the nearest ten would be 60 pounds.

Meanwhile, Jane is 67 pounds. 7 is more than 5, so we will raise the score.

So, 67 rounded to the nearest ten would be 70.

You have your answers. 60 and 70. Hope this helps!

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Answer:

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Step-by-step explanation:

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2 years ago
Suppose the selling price of homes is skewed right with a mean of 350,000 and a standard deviation of 160000 If we record the se
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Answer:

The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

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Population:

Suppose the selling price of homes is skewed right with a mean of 350,000 and a standard deviation of 160000

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Shape approximately normal

Mean 350000

Standard deviation s = \frac{160000}{\sqrt{40}} = 25298

The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.

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Step-by-step explanation:

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