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Fynjy0 [20]
3 years ago
14

A hot air balloon gets loose from its anchor the balloon operator is look at the balloon at an angle of inclination of 34. If th

e balloon is 50 feet in the air, what is the horizontal distance from the balloon to the balloon operator
Mathematics
1 answer:
masha68 [24]3 years ago
4 0

Answer:

horizontal distance = 74.13 ft

Step-by-step explanation:

In order to solve this problem, we can start by drawing what the situation looks like. (see attached picture)

As we can see in the picture, we can build a right triangle with the balloon and the operator. We can use a trigonometric function so solve this. Specifically the tan function. We know that:

tan \theta = \frac{opposite}{adjascent}

in this case:

tan \theta = \frac{h}{l}

so we need to figure out what l is equal to. We can solve this equation for l, so we get:

l=\frac{h}{tan \theta}

and substitute the values given by the problem, so we get:

l=\frac{50ft}{tan (30^{0})}

which yields:

l=74.13 ft

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Answer:

Step-by-step explanation:

56.1

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3 years ago
The product of 110 and h, plus 115 equals 63
Dima020 [189]

Answer:

h = -0.5

Step-by-step explanation:

110h + 115 = 63

63 - 115

= -52

110h = -52

h = -0.5

7 0
3 years ago
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Y=-7x+4<br> Y=3/8x+-4<br> Y=16x+-1/5
zavuch27 [327]

Answer:

  1. y = -7x +4
  2. y = 3/8x - 3
  3. y = 16x -1/5

5 0
4 years ago
A population is estimated to have a standard deviation of 9. We want to estimate the population mean within 2, with a 99% level
cupoosta [38]

Answer:

The sample required is  n = 135

Step-by-step explanation:

From the question we are told that

     The  standard deviation is  \sigma = 9

      The margin of error is E =  2

     

Given that the confidence level is  99%  then the level of  significance is mathematically evaluated as

         \alpha =  100-99

        \alpha =  1 \%

        \alpha =  0.01

Next we will obtain the critical value  \frac{\alpha }{2} from the normal distribution table(reference  math dot armstrong dot edu) , the value is  

             Z_{\frac{\alpha }{2} } =  Z_{\frac{0.05 }{2} } =  2.58

The  sample size is mathematically represented as

          n = [ \frac{Z_{\frac{\alpha }{2} } *  \sigma }{E} ]^2

substituting values

           n = [ \frac{ 2.58 *  9 }{2} ]^2

            n = 135

3 0
3 years ago
In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 60
Olegator [25]

Answer:

7.38-1.96\frac{2.51}{\sqrt{601}}=7.18    

7.38+1.96\frac{2.51}{\sqrt{601}}=7.58    

For this case the confidence interval is given by :

7.18 \leq \mu \leq 7.58

Step-by-step explanation:

Data provided

\bar X=7.38 represent the sample mean for the fuel efficiencies

\mu population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size  

Confidence interval

The formula for the confidence interval of the true mean is given by:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom for this case is given by:

df=n-1=601-1=600

The Confidence level for this case is 0.95 or 95%, and the significance level \alpha=0.05 and \alpha/2 =0.025 and the critical value is given by t_{\alpha/2}=1.96

Replcing into the formula for the confidence interval is given by:

7.38-1.96\frac{2.51}{\sqrt{601}}=7.18    

7.38+1.96\frac{2.51}{\sqrt{601}}=7.58    

For this case the confidence interval is given by :

7.18 \leq \mu \leq 7.58

8 0
4 years ago
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