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adelina 88 [10]
3 years ago
15

A portion of a line or Ray that extends from one point to another point is called what

Mathematics
2 answers:
madam [21]3 years ago
6 0

Answer:

line segment

Step-by-step explanation:

If a portion of a line has two endpoints (two little black dots on its end), then it is a line <em>segment</em>.

If it has one dot and an arrow, its a <em>ray</em>. This has an endpoint and goes on forever in the direction of the arrow.

Juli2301 [7.4K]3 years ago
5 0

Answer: line segment

Step-by-step explanation:

A line segment is a ray that extends from one point to another. Hope this helps!

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Solve the linear equation x + 4 + 3 + 72
elixir [45]

Answer:

x=17

Step-by-step explanation

4x+4=72

-4      -4

4x=68

/4 /4

x= 17

7 0
2 years ago
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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

8 0
3 years ago
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cluponka [151]
The answer is B.98 
180-122=58
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3 years ago
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Mariana [72]

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2 years ago
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