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V125BC [204]
3 years ago
11

Three pounds of fertilizer covers 900 feet.How many pound of fertilizer will cover 2,100 feet?

Mathematics
1 answer:
sveticcg [70]3 years ago
7 0
To solve this problem, you can make a proportion:

3/900 = x/2100

In this case, the x represents the pounds that will be needed to cover 2,100 feet.

The first step is to cross multiply like this:

3 * 2100 = 900 * x

Now you can simply the equation.

6300 = 900x

Now you divide both sides by 900 to cancel it out.

7 = x

So, your answer is 7 pounds.
You might be interested in
In the equation 3s +135= 4.5 (s +10) what is s
Lubov Fominskaja [6]
3s + 135 = 4.5s + 45

Now move the variables and coefficients

135 = 1.5s + 45

And then the numbers

90 = 1.5s

Simplify

90/1.5 = 1.5s/1.5

60 = s

Hope this helps!
6 0
2 years ago
It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So n = 6, s = \frac{68}{\sqrt{6}} = 27.76

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 375}{27.76}

Z = 1.26

Z = 1.26 has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So \mu = 522, \sigma = 106

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So n = 6, s = \frac{106}{\sqrt{6}} = 43.27

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 523}{43.27}

Z = -2.61

Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0
3 years ago
I need urgent help with this one
Serhud [2]

let the unknown answer be equal to f

.-2.1 - f = -3/2

-f = -3/2 + 2.1

-f = 0.6

f = -0.6

5 0
2 years ago
Read 2 more answers
I'll give brainliest and five stars if you help me!
andrezito [222]
A.)))))))))))


- Harvard university professor
6 0
3 years ago
A dealership paid $8,000 for a car at an
sertanlavr [38]
3,200 is 40% of 8000 although im confused on what ‘157’ commission is?
8 0
3 years ago
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