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Korolek [52]
2 years ago
13

Miles takes a 350-mile round-trip flight to visit his parents. To qualify for Gold status at Awesome

Mathematics
1 answer:
Vika [28.1K]2 years ago
6 0

Answer:

Miles must travel a minimum of 22 times and a maximum of 36 times to achieve Gold status.

Step-by-step explanation:

Given that Miles takes a 350-mile round-trip flight to visit his parents, and to qualify for Gold status at Awesome Airlines, one must fly at least 7700 and less than 12600 miles each year, to determine how many times would Miles need to visit his parents each year to attain Gold status, the following calculation must be performed:

7700/350 = X

22 = X

12600/350 = X

36 = X

Thus, Miles must travel a minimum of 22 times and a maximum of 36 times to achieve Gold status.

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How do you simplify a fraction?
attashe74 [19]
You simplify a fraction by finding it's greatest common factor. A greatest common factor is what both numbers are divisible by. If they can both be divided by two, then you've simplified them a little bit. Keep going until you can't find a common number that can be divided by the two, and you've got your simplified fraction. For example, 10/5 = 1/2 since both 10 and 5 is divisible by 5.
5 0
3 years ago
Scores on a final exam taken by 1200 students have a bell shaped distribution with mean=72 and standard deviation=9
SVETLANKA909090 [29]

Answer:

a. 72

b. 816

c. 570

d. 30

Step-by-step explanation:

Given the graph is a bell - shaped curve. So, we understand that this is a normal distribution and that the bell - shaped curve is a symmetric curve.

Please refer the figure for a better understanding.

a. In a normal distribution, Mean = Median = Mode

Therefore, Median = Mean = 72

b. We have to know that 68% of the values are within the first standard deviation of the mean.

i.e., 68% values are between Mean $ \pm $ Standard Deviation (SD).

Scores between 63 and 81 :

Note that 72 - 9 = 63 and

72 + 9 = 81

This implies scores between 63 and 81 constitute 68% of the values, 34% each, since the curve is symmetric.

Now, Scores between 63 and 81 = $ \frac{68}{100} \times 1200 $

= 68 X 12 = 816.

That means 816 students have scored between 63 and 81.

c. We have to know that 95% of the values lie between second Standard Deviation of the mean.

i.e., 95% values are between Mean $ \pm $ 2(SD).

Note that 90 = 72 + 2(9) = 72 + 18

Also, 54 = 63 - 18.

Scores between 54 and 90 totally constitute 95% of the values. So, Scores between 72 and 90 should amount to $ \frac{95}{2} \% $ of the values.

Therefore, Scores between 72 and 90 = $ \frac{95}{2(100)} \times 1200 = \frac{95}{200} \times 1200  $

$ \implies 95 \times 12 $ = 570.

That is a total of 570 students scored between 72 and 90.

d. We have to know that 5 % of the values lie on the thirst standard Deviation of the mean.

In this case, 5 % of the values lie between below 54 and above 90.

Since, we are asked to find scores below 54. It should be 2.5% of the values.

So, Scores below 54 = $ \frac{2.5}{100} \times 1200 $

= 2.5 X 12 = 30.

That is, 30 students have scored below 54.

8 0
2 years ago
Positive , negative or zero
Bess [88]

Answer:

negative, it'll be negative 2

3 0
2 years ago
Suppose that the functions g and h are defined for all real numbers r as follows. gx) -4x +5 h (x) = 6x write the expressions fo
aleksandr82 [10.1K]

Answer: Our required values would be -10x+5, 2x+5 and -25.

Step-by-step explanation:

Since we have given that

g(x) = -4x+5

and

h(x) = 6x

We need to find  (g-h)(x) and (g+h)(x).

So, (g-h)(x) is given by

g(x)-h(x)\\\\=-4x+5-6x\\\\=-10x+5

and (g+h)(x) is given by

g(x)+h(x)\\\\=-4x+5+6x\\\\=2x+5

and (g-h)(3) is given by

-10(3)+5\\\\=-30+5\\\\=-25

Hence, our required values would be -10x+5, 2x+5 and -25.

5 0
3 years ago
PLZ HELP ME QUICK AT THIS MATH PROBLEM QUICK
sukhopar [10]

Answer:

1. 30 houses were analyzed

2. 24 houses

3. 21 houses

4. Bimodal skewed right? IDK sorry :(

5. 15 houses

6. The fourth interval (151-200)

Sorry if this is wrong... Hope it helps!

Step-by-step explanation:

4 0
2 years ago
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