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BaLLatris [955]
3 years ago
9

2= 5 j -28 one or two step equation

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

j=6

Step-by-step explanation:

2=5j-28

+28  +28

30=5j

/5    /5

j=6  

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Dylan works 30 hours a week. His hours are decreased by 20%. How many hours does he now work each week?
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20% of 30 is 6. 30-6 is 24. He now works for 24 hours! Please mark brainliest!!!
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Find the simple interest on 500 at 6% for 1 year
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Answer:

Interest = $30

Step-by-step explanation:

Formula:

I = prt or I = p · r · t

So:

I = p r t

I = ($500)(.06)(1) <--- Just multiply all of these numbers and you get the final answer of:

Interest = $30.00 or just $30

8 0
3 years ago
Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1
svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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 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.

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.

Mean of 8.0 ounces and a standard deviation of 1.1 ounces.

This means that \mu = 8, \sigma = 1.1

(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?

n = 5 means that s = \frac{1.1}{\sqrt{5}} = 0.4919

This probability is the pvalue of Z when X = 9.3. So

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

By the Central Limit Theorem

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

Z = \frac{9.3 - 8}{0.4919}

Z = 2.64

Z = 2.64 has a pvalue of 0.9959

0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces?

n = 6 means that s = \frac{1.1}{\sqrt{6}} = 0.4491

This probability is 1 subtracted by the pvalue of Z when X = 9. So

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

Z = \frac{9 - 8}{0.4491}

Z = 2.23

Z = 2.23 has a pvalue of 0.9871

1 - 0.9871 = 0.0129

0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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Step-by-step explanation:

The alternate interior angles look like the picture provided below. Next time, don't clickbait with the wrong amount of points.

QPS is included but isn't one of the answers because they it's not a multiple choice question.

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