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2 years ago

Which of the following appear in the diagram below??

Mathematics

2 answers:

Alina [70]2 years ago

8 0

Answer:

The correct answer are A, B and C.

Step-by-step explanation:

The line through the AB points is clearly defined at the top of the graph.

The DCE segment appears below the graph forming an angle by the union of these three points.

The CE segment is a vector marked in the same DCE set, only it does not contain point D.

Finally, the CDE segment is not possible, as there is no common line between points D and E.

Have a nice day!

insens350 [35]2 years ago

7 0

The ones that appear are A B and C

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A monkey has filled in a 3 x 3 grid with the numbers 1, 2,...,9. A cat writes down the three numbers obtained by multiplying the

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Answer:

Deatailed Answer to the question is given in the attached files.

Step-by-step explanation:

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2 years ago

Katie wants to create a rectangular frame for a picture. She has 60 inches of material. If she wants the length to be 3 more tha

Fofino [41]

Answer:

Largest possible length is <em>21 inches</em>.

Step-by-step explanation:

Given:

Total material available = 60 inches

Length to be 3 more than twice of width.

To find:

Largest possible length = ?

Solution:

As it is rectangular shaped frame.

Let length = $l$ inches and

Width = $w$ inches

As per given condition:

$l = 2w+3$ ..... (1)

Total frame available = 60 inches.

i.e. it will be the perimeter of the rectangle.

Formula for perimeter of rectangle is given as:

$P = 2 \times (Width + Length)$

Putting the given values and conditions as per equation (1):

$60 = 2 \times (w+ l)\\\Rightarrow 60 = 2 \times (w+ 2w+3)\\\Rightarrow 60 = 2 \times (3w+3)\\\Rightarrow 30 = 3w+3\\\Rightarrow 3w = 27\\\Rightarrow w = 9 \ inch$

Putting in equation (1):

$l = 2\times 9+3\\\Rightarrow l = 21\ inch$

So, the answer is:

Largest possible length is <em>21 inches</em>.

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2 years ago

A greengrocer sells a melon at a profit of 37.5% on the price he pays for it. what is the ratio the cost price to the selling pr

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2 years ago

A manager at a local manufacturing company has been monitoring the output of one of the machines used to manufacture chromium sh

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Answer:

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

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.

Normally distributed with a mean of 118 centimeters and a standard deviation of 8 centimeters.

This means that $\mu = 118, \sigma = 8$

Sample of 16 shells

This means that $n = 16, s = \frac{8}{\sqrt{16}} = 2$

What is the probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly?"

This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 116.

X = 120

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

By the Central Limit Theorem

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

$Z = \frac{120 - 118}{2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.841

X = 116

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

$Z = \frac{116 - 118}{2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.159

0.841 - 0.159 = 0.682

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

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