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sdas [7]
3 years ago
13

HELP PLEASE ((ATTACHMENT TO HELP ANSWER, BECAUSE THERE'S REALLY NO QUESTION WITH IT))

Mathematics
1 answer:
BabaBlast [244]3 years ago
7 0
Linear- blue graph, table 2, 2x-3y=7, y=3\x +2, y=-4
Nonlinear- y= 2x^2+ 4
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How do you write 6x=7y in standard form
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Standard form is ax+by=c
mnus 7y both sides
6x-7y=0
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Mr Coope drives 472 miles in 8 hours. At this rate, how many miles would he drive in 6
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8hrs = 480mins
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472/480 = 0.983...
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D) 354
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What is 102.75609756 as a whole number please answer
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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
Solve for z -9(d+z)=-2z+59
neonofarm [45]
-9(d+z)=-2z+59\\\\\text{Use multiply distributive property:}\\a\cdot(b+c)=ab+ac\\\\-9d-9z=-2z+59\ \ \ |+9d\\\\-9z=-2z+59+9d\ \ \ |+2z\\\\-7z=59+9d\ \ \ \ |:(-7)\\\\\boxed{z=-\dfrac{59+9d}{7}}

7 0
3 years ago
Read 2 more answers
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