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frutty [35]
2 years ago
15

Determine the domain and the range for this relationship.

Mathematics
1 answer:
Vladimir [108]2 years ago
7 0

Answer:

domain 0 ≤ y ≤ 72 and the range is 0 ≤ y ≤ 6

Step-by-step explanation:

I hope this helps you.

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F(x) = -2(x + 2)2 + 4
USPshnik [31]

Answer:

4x4=?        then we have 2x2=?

Step-by-step explanation: H-JJ-

7 0
2 years ago
I need help in the question 7
Alona [7]
6.75 in by 5 in. the area is 33.75
4 0
3 years ago
Nikita ran a 5-kilometer race in 39 minutes (0.65 of an hour) without training beforehand. In the first part of the race, her av
s344n2d4d5 [400]

Nikita's distance from her for the second part of the race was 1.5 kilometers.

Since Nikita ran a 5-kilometer race in 39 minutes (0.65 of an hour) without training beforehand, and in the first part of the race, her average speed was 8.75 kilometers per hour, while for the second part of the race, she started to get tired, so her average speed dropped to 6 kilometers per hour, to determine which expression represents Nikita's distance for the second part of the race the following calculation must be performed:

  • 8.75 x 0.65 + 6 x 0 = 5.68
  • 8.75 x 0.50 + 6 x 0.15 = 5.275
  • 8.75 x 0.40 + 6 x 0.25 = 5
  • 100 = 60
  • 40 = X
  • 40 x 60/100 = X
  • 24 = X
  • 39 - 24 = 15
  • 15 = 1/4 hour
  • 6/4 = 1.5

Therefore, Nikita's distance from her for the second part of the race was 1.5 kilometers.

Learn more about maths in brainly.com/question/22791791

6 0
2 years ago
in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru
melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is  

n = 200

the proportion is

p = 154/200 = 0.77

<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>

<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant  difference. </em>

The approximation would be to a Normal curve with this parameters:

<em>Mean </em>

p = 0.77

<em>Standard deviation </em>

\bf s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.77*0.23}{200}}=0.0298

The 90% confidence interval for the proportion would be then  

\bf  [0.77-z^*0.0298, 0.77+z^*0.0298]

where \bf z^* is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval \bf [-z^*,z^*] is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

<em>NORMSINV(0.95) and we get a value of 1.645 </em>

The 90% confidence interval for the proportion is then

\bf  [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

5 0
3 years ago
If my birthday is September 1st 2007 how old am I
Lelechka [254]
You are 13 years old
7 0
3 years ago
Read 2 more answers
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