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andriy [413]
3 years ago
6

A Euler circuit has ___ ODD vertices

Mathematics
1 answer:
xenn [34]3 years ago
4 0

Answer:

A Euler circuit has TWO ODD vertices

Step-by-step explanation:

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17. Atlanta, Georgia, receives an average of 27 inches of precipitation per year, and there has been 9 inches so far this year.
kirill115 [55]

Answer:

  9 + p ≤ 27; p ≤ 18

Step-by-step explanation:

"At or below" means "less than or equal to." Only one answer choice incorporatest that symbol into the inequality expression — the one shown above.

5 0
2 years ago
Read 2 more answers
8 x - 4 (5 - x) = -44
Sphinxa [80]

8 x - 4 (5 - x) = -44

mutiply the bracket by -4

(-4)(5) = -20

(-4)(-x)= 4x

8x-20+4x= -44

8x+4x-20= -44 ( combine like terms )

12x-20= -44

move -20 to the other side

sign changes from -20 to +20

12x-20+20= -44+20

12x= -44+20

12x= -24

divide both sides by 12

12x/12= -24/12

Answer: x= -2

3 0
3 years ago
Answer Is not C<br>please help​
Katyanochek1 [597]

Answer:

B

Step-by-step explanation:

Well since it is arthimetic sequence we know the formula

We know d is difference between second term and first so it will be -5- (-12)= -5+12= 7

a_{n}= a _{1}+ d (n-1) \\a_{n}= -12+7(n-1)\\a_{n} = -12+7n-7\\a_{n}= 7n-19

the domain of this is 1,2,3,4.......

range is -12,-5,2,9.......

Therefore answer is B.

5 0
3 years ago
One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla
Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

6 0
3 years ago
Jason and Britton are driving to St. George. Jason got a 20 mile head start and drove an
harina [27]

Answer:

  2 hours, 150 miles

Step-by-step explanation:

The relation between time, speed, and distance can be used to solve this problem. It can work well to consider just the distance between the drivers, and the speed at which that is changing.

<h3>Separation distance</h3>

Jason got a head start of 20 miles, so that is the initial separation between the two drivers.

<h3>Closure speed</h3>

Jason is driving 10 mph faster than Britton, so is closing the initial separation gap at that rate.

<h3>Closure time</h3>

The relevant relation is ...

  time = distance/speed

Then the time it takes to reduce the separation distance to zero is ...

  closure time = separation distance / closure speed = 20 mi / (10 mi/h)

  closure time = 2 h

Britton will catch up to Jason after 2 hours. In that time, Britton will have driven (2 h)(75 mi/h) = 150 miles.

__

<em>Additional comment</em>

The attached graph shows the distance driven as a function of time from when Britton started. The distances will be equal after 2 hours, meaning the drivers are in the same place, 150 miles from their starting spot.

3 0
1 year ago
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