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Mamont248 [21]
3 years ago
5

In the figure below, C is between A and D, and B is the midpoint of AC. If AD=20 and BC=6, find CD.

Mathematics
2 answers:
Leokris [45]3 years ago
8 0

Answer:

8

Step-by-step explanation:

AC and BC are the same, so both equal 6

6+6 is 12

Subtract 12 from 20

20-12=8

CD is 8

Vesna [10]3 years ago
8 0
8 i think, sorry if its wrong:) I can give an explanation if needed!
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Finger [1]

Answer:

17, 558 employees

Step-by-step explanation:

Altogether = 70,010 employees , First location = 34, 857 employees, second location = 17, 595 employees, Third location = ?

Let the number of employees in the third location be x

70,010 + 17, 595 + x = 70,010

x = 70, 010 - 52452

x = 17, 558( Answer)

17, 558 employees work in the third location

3 0
2 years ago
My number is a multiple of 2 & 7 my number is less than 100 but greater than 50 my number is the product of three prime numb
forsale [732]
12345678910
ihtrguhbvr;wmytpb jq
omuyi bnkq4ht6kljihtlr'iwhvncnh ug5

8 0
2 years ago
What's the greatest common factor of 5x^2+3x
schepotkina [342]
The answer to that question is (5x+3).
6 0
3 years ago
Read 2 more answers
PLEASE HELP ME I BEG I WILL GIVE BRAINLIEST
ohaa [14]

Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

Step-by-step explanation:

7 0
3 years ago
Rob notices that 55 percent of the people leaving the supermarket chhose plastic bags instead of paper bags. out of 600 people,
Temka [501]
To solve this answer, you are trying to find 55% of 600, because that is the number of people who chose plastic bag.

55% can be re-written as 55/100 OR 0.55 since percentages are always out of 100.

So, we do 0.55*600, which gives us 330. 330 people carry plastic bags.
4 0
3 years ago
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