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3 years ago

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Drag and drop all methods of showing Triangle Similarity or Triangle Congruence into the appropriate box. SAS, SSS, ASA, AAS, CP

OCTAC, HL, AA
Similar Triangle Justifications Congruent Triangle Justifications

1 answer:

Maksim231197 [3]3 years ago

5 0

Answer:

hi

Step-by-step explanation:

by

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Due to the increase of posts on social media, the number of flip cell phones is in exponential decline. Data collected from 2009

gizmo_the_mogwai [7]

Answer:

$y=256e^{-0.288x}$

where y is in thousands

Step-by-step explanation:

The equation in two variables that can be used to represent the situation given can be determined using computer software such Stat-Crunch or Ms. Excel or advanced calculators.

I shall use Ms. Excel for this question.

In Ms. Excel, enter the data into any two adjacent columns. Next, highlight the data, then click the insert ribbon and select the scatter-plot option.

Excel returns a scatter-plot chart as shown in the attachment below.

After obtaining the scatter-plot, we shall need to add a trend line in order to determine the mathematical equation that best fits the scenario given.

Click anywhere inside the chart, then select the design tab under chart tools. Click on the Add Chart element in the upper left corner of the excel workbook and select more trend-line options. This feature will enable us to fit any trend-line to our data.

Select the exponential option, since we were informed that he number of flip cell phones is in exponential decline,ensuring you check the box; Display Equation on chart.

Find the attached.

The equation in two variables is thus;

$y=256e^{-0.288x}$

5 0

3 years ago

Solve f(1) for f(x)=1-5x<br><br> f(1)= [?]

Tpy6a [65]

The answer is: "- 4 " .
________________________________________
" f(x) = - 4 " .
________________________________________
Explanation:
________________________________________
Given: " f(x) = 1 – 5x " ; Find: " f(1) " ;

Substitute "1" for values of "x" ;

→ f(1) = 1 – 5(1) = 1 – 5 = - 4 .

The answer is: "- 4 " .
________________________________________
" f(x) = - 4 " .
________________________________________

5 0

3 years ago

A. A batch of 30 parts contains five defects. If two parts are drawn randomly one at a time without replacement, what is the pro

Zarrin [17]

Answer:

a) For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

For the second trial since the experiment is without replacement we have 29 parts left, and since we select 1 defective from the first trial we have in total 4 defective left, so then the probability of defective for the second trial is : 4/29

And then we can assume independence between the events and we have this probability:

(5/30)*(4/29) =0.16667*0.1379= 0.0230

b) For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

And since we replace the part selected is the same probability for the second trial and then the final probability assuming independence would be:

(5/30)*(5/30) =0.1667* 0.1667= 0.0278

Step-by-step explanation:

For this case we know that we have a batch of 30 parts with 5 defective.

Part a

If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?

For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

For the second trial since the experiment is without replacement we have 29 parts left, and since we select 1 defective from the first trial we have in total 4 defective left, so then the probability of defective for the second trial is : 4/29

And then we can assume independence between the events and we have this probability:

(5/30)*(4/29) =0.16667*0.1379= 0.0230

Part b

If this experiment is repeated, with replacement, what is the probability that both parts are defective?

For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

And since we replace the part selected is the same probability for the second trial and then the final probability assuming independence would be:

(5/30)*(5/30) =0.1667* 0.1667= 0.0278

3 0

3 years ago

How many times can 10 go into 111? lol i need help

ikadub [295]

Answer:

11.1 times

Step-by-step explanation:

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2 years ago

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This mapping shows a functional relationship.

Dmitriy789 [7]

Answer:

Step-by-step explanation:

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3 years ago