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RUDIKE [14]
3 years ago
6

So,i have the answer to this question,but i have no clue how to work it out,ik im not that smart but anyways i need helppo

Mathematics
1 answer:
mestny [16]3 years ago
7 0

Answer:

1290 =  \frac{h}{10}  + \frac{h}{5} \\ 12900 = h + 2h \\ 3h = 12900 \\ h = 4300

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Explain why |−3| + |9| represents the distance between the points (−3, −5) and (9, −5). brainly
forsale [732]
Since -5 and -5 are the same number, they are on the same y axis so the only distance we need to calculate is between the X coordinates.

When looking for distance, you can not have a negative. what |-3|+|9| is looking for is the distance from 0 on the X axis. since distance can't be negative, (you go 3 miles away from your house in one direction, vs 3 miles in the opposite. both ways is positive, even with opposite directions. same with axis, it doesnt matter which way, only the number.) you need the absolute value of -3 to get the distance from the 0 on the x axis. 

short version: distance is positive, and its adding the distances from the x axis to get distance from each other
8 0
3 years ago
Read 2 more answers
The possible values of x are given by x−3≥2 . What is the least possible value of 5x ?
mamaluj [8]
The answer would be twenty five because x-3 \geq 2
 equals x  \geq 5 leaving you with x = 5 at a minimum. So 5x: 5 x 5 = 25

Hope this helps! :)
5 0
3 years ago
When jack went to the movie he paid $8 for his popcorn and $4 for each additional order of popcorn. if i represents the total co
melamori03 [73]
I = 8 + 4x

XD That's expensive popcorn!
8 0
3 years ago
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A factory dumps an average of 2.43 tons of pollutants into a river every week. if the standard deviation is 0.88 tons, what is t
vovikov84 [41]
25.8%  
First, determine how many standard deviations from the norm that 3 tons are. So:
 (3 - 2.43) / 0.88 = 0.57/0.88 = 0.647727273  
So 3 tons would be 0.647727273 deviations from the norm. Now using a standard normal table, lookup the value 0.65 (the table I'm using has z-values to only 2 decimal places, so I rounded the z-value I got from 0.647727273 to 0.65). The value I got is 0.24215. Now this value is the probability of getting a value between the mean and the z-score. What I want is the probability of getting that z-score and anything higher. So subtract the value from 0.5, so 0.5 - 0.24215 = 0.25785 = 25.785%  
So the probability that more than 3 tons will be dumped in a week is 25.8%
5 0
3 years ago
The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili
Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that \mu = 74, \sigma = 15

Question a:

Sample of 36 means that n = 36, s = \frac{15}{\sqrt{36}} = 2.5

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

By the Central Limit Theorem

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

Z = \frac{78 - 74}{2.5}

Z = 1.6

Z = 1.6 has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that n = 150, s = \frac{15}{\sqrt{150}} = 1.2247

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

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

Z = \frac{77 - 74}{1.2274}

Z = 2.45

Z = 2.45 has a pvalue of 0.9929

X = 71

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

Z = \frac{71 - 74}{1.2274}

Z = -2.45

Z = -2.45 has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that n = 219, s = \frac{15}{\sqrt{219}} = 1.0136

This probability is the pvalue of Z when X = 74.2. So

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

Z = \frac{74.2 - 74}{1.0136}

Z = 0.2

Z = 0.2 has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0
3 years ago
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