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vovangra [49]
3 years ago
12

Given that Line segment B C is parallel to line segment E F which theorem or postulate proves ΔBCD =ΔFED

Mathematics
1 answer:
raketka [301]3 years ago
7 0
Seems like a trick question to me. Unless it states that D bisects CE and BF, then you can't say for sure.
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A person suspects a standard number cube is not so standard. Another person rolls this cube 100 times, and it lands on a six 42
Ksju [112]

Answer:

The chance of an average six-sided number cube landing one one number is 1/6, or approximately 16.67%. When rolling a number cube 100 times, the average number of times it lands on any specific number should be fairly close to 16.67.

A 6 being rolled 42 times and 33 times means that it is <em>very likely</em> not to be a standard number cube. Although we cannot know for certain, the chance of this happening is very low.

3 0
3 years ago
If you have the fraction 8/25, and you are trying to turn it into a decimal, which of the following would be helpful?
AysviL [449]
In this situation you will need to do 8 divided by 25 which will give the answer 0.32.
8 0
3 years ago
Read 2 more answers
Select all values of x that make the inequality -x + 8 ≥11 true.
mash [69]

-x + 8 ≥ 11

-x ≥ 3

x ≤ -3

Hope this helps! ;)

8 0
3 years ago
Of those women who are diagnosed to have early-stage breast cancer, one-third eventually die of the disease. Suppose a screening
Finger [1]

Answer:

A) Alternative hypothesis; Ha: p_o > ⅔

B) Null hypothesis;

H0: p_o = ⅔

C) There is sufficient evidence to support the claim that the community screening programme was effective

D) p-value = 0

Its less than the significance value and so we will reject the null hypothesis and conclude that the community screening programme was effective

Step-by-step explanation:

A) We are told that ⅓ of those diagnosed eventually die of the disease. Thus, ⅔ is the fraction that will survive the disease.

Thus, the null hypothesis is;

H0: p_o = ⅔

We are told that a screening programme was started to increase the survival rate. Thus,the alternative hypothesis is;

Ha: p_o > ⅔

B) null hypothesis is;

H0: p_o = ⅔

C) 164 out of the 200 women selected survived the disease.

Thus sample proportion is;

p^ = 164/200

p^ = 0.82

For the z-score value, we will use the formula;

z = (p^ - p_o)/√((p_o(1 - p_o)/n)

Now, p_o = ⅔ = 0.67

Thus;

z = (0.82 - 0.67)/√((0.67(1 - 0.67)/200)

z = 4.51

From online p-value from z-score calculator, we have p-value ≈ 0

This is less than the significant level of 0.05 and therefore we will reject the null hypothesis and conclude that the community screening program was effective.

D) p-value ≈ 0

Reject the null hypothesis

4 0
2 years ago
What is the first quartile of the data displayed in this box-and-whisker plot?
ValentinkaMS [17]
4. 42,


Since Q1 is the first part of the box plot, and on the graph states
42
7 0
3 years ago
Read 2 more answers
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