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Degger [83]
2 years ago
7

In the triangle below, what is the length of EF A. 6 B. 24 C. 12 D. 25

Mathematics
1 answer:
Ymorist [56]2 years ago
6 0

Answer:

12

Step-by-step explanation:

since DE and EF are congruent then we now know that EF is also 12

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The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population o
blondinia [14]

The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

According to the statement

we have to find that the standard deviation, mean and the intervals from the given data.

So, According to the given data from the method of tree ring dating

The value of mean is

x-bar = (1271 + 1208 + 1229 + 1299 + 1268 + 1316 + 1275 + 1317 + 1275) / 9 = x-bar = 1273 AD

And now we find standard deviation :

s = √∑(xi - x-bar) / (N - 1)

∑(xi - x-bar)^2 = (1271 - 1273)2 + (1208 - 1273)2 + (1229 - 1273)2 + ... + (1275 - 1273)2

∑(xi - x-bar)^2 = (-2)2 + (-65)2 + (-44)2 + ... + (2)2

∑(xi - x-bar)^2 = 4 + 4225 + 1936 + 676 + 25 + 1849 + 4 1936 + 4

∑(xi - x-bar)^2 = 10,659

Now,

s^2 = 10659/8 = 1332

s = 37 years

So, standard deviation is 37 years.

We need the t-distribution table since the standard deviation is unknown.  Therefore, our degrees of freedom is 9 - 1 = 8 and the critical value is 1.860.  Set up the confidence interval for the mean:

[x-bar ± t*(s/√n)] = [1273 ± 1.860*(37/√9)]

[x-bar ± t*(s/√n)] = [1250,1296]

So, The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

Learn more about method of tree ring dating here

brainly.com/question/15107034

Disclaimer: This question was incomplete. Please find the full content below.

Question:

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.

For more data please see the image below.

#SPJ4

4 0
1 year ago
Dalton when to high school at 11 yrs and 3 months he left high school at 18 yrs and 2 months old how long did Dalton spend in hi
Tanya [424]

Answer:

Dalton spent 6 years and 11 months in high school.

Step-by-step explanation:

Dalton were in high school at the age = 11 years and 3 months                       He left high school at the age = 18 years and 2 months

Time spent by Dalton in high school = 18 years 2 months - 11 years and 3 months

= (18 - 11) years and (2 - 3) months

= 7 years - 1 month

= (6 + 1) years - 1 month

= 6 years + 1 year - 1 month

= 6years + 12 months - 1 month

= 6 years + 11 months

Therefore, Dalton spent 6 years and 11 months in high school.

6 0
2 years ago
A bag contains 6 green counters, 4 blue counters and 2 red counters. Two counters are drawn from the bag at random without repla
Svetach [21]

Answer:

24.24%

Step-by-step explanation:

In other words we need to find the probability of getting one blue counter and another non-blue counter in the two picks. Based on the stats provided, there are a total of 12 counters (6 + 4 + 2), out of which only 4 are blue. This means that the probability for the first counter chosen being blue is 4/12

Since we do not replace the counter, we now have a total of 11 counters. Since the second counter cannot be blue, then we have 8 possible choices. This means that the probability of the second counter not being blue is 8/11. Now we need to multiply these two probabilities together to calculate the probability of choosing only one blue counter and one non-blue counter in two picks.

\frac{4}{12} * \frac{8}{11} =   \frac{32}{132} or 0.2424 or 24.24%

8 0
3 years ago
What is the function, domain, and range??<br> {(1,-2), (-2,0), (-1,2), (1,3)}
scoundrel [369]

Answer:

the relation is NOT a function

domain: { 1,-2,1}

range:  { -2, 0, 2, 3}

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What's the answer for each to get 30 points!
MAVERICK [17]

Answer:  "greater than" for each of the 4 dropdown menus

======================================================

Explanation:

Divide each value in the table by 40

You should get:

  • Horse = 15/40 = 0.375
  • Cow = 12/40 = 0.300
  • Sheep = 14/40 = 0.350
  • Pig = 19/40 = 0.475

Those decimal results are the experimental (ie empirical) probabilities. Theoretically, we should get 1/4 = 0.250 for each sticker type assuming each sticker is likely to be chosen. As you can see, each decimal value shown above is larger than the theoretical target of 0.250, so each answer is "greater than"

-----------

Here's another way to see why this is:

If we had 40 stickers total, and each animal has the same number of stickers, then we should have 40/4 = 10 stickers per animal type. But the table shows each frequency is above 10. So that must directly mean the empirical probability of picking any animal is greater than the theoretical probability.

6 0
3 years ago
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