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DanielleElmas [232]
3 years ago
7

Some farms use a hay elevator to move bales of hay to the second story of a barn loft. The bottom of the elevator is 9 meters fr

om the barn and the elevator makes an angle of 34° with the ground. What is the height of the barn from the bottom of the barn opening to the ground?

Mathematics
2 answers:
Serjik [45]3 years ago
5 0

Answer:

6.071 metres

Step-by-step explanation:

Please kindly see the attached file for more explanation

-Dominant- [34]3 years ago
3 0

Answer: The height is 6.07 meters

Step-by-step Explanation: The distance between the elevator and the bottom of the barn is given as 9 meters. Also for the hay elevator to move bales of hay to the second story of the barn lift it makes an angle of elevation of 34 degrees with the ground. With these we can derive a right angled triangle with the reference angle as 34 degrees, the side facing it which is the height or h (opposite) is yet unknown, and the side between the reference angle and the right angle (adjacent) is 9. We shall apply the trigonometric ratio as follows;

Tan 34 = opposite/adjacent

Tan 34 = h/9

0.6745 = h/9

0.6745 x 9 = h

6.0705 = h

Therefore the approximate height of the barn to the ground is 6.07 meters

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Natasha2012 [34]

Answer:

10=19-11

Step-by-step explanation:

that answer is false because 19-11 = 8 not 10

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3 years ago
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miss Akunina [59]

Answer:

  2.5 miles

Step-by-step explanation:

The relation between time, speed, and distance is ...

  distance = speed × time

We can define t to be Stanley's swimming time. Then t+0.5 was his running time, and 2(t+0.5) was his biking time. His total distance covered is ...

  64 = 9(t +0.5) +16(2(t +0.5)) +2.5(t)

  64 = 43.5t +20.5 . . . . . . . simplify

  43.5 = 43.5t . . . . . . . . . subtract 20.5

  t = 1 . . . . . . . . . . . . . . divide by the coefficient of t

Stanley swam for 1 hour, so the distance he covered while swimming was ...

  (2.5 mi/h)(1 h) = 2.5 mi

Stanley covered 2.5 miles while swimming.

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<em>Additional comment</em>

Stanley ran for 1.5 hours, covering 9×1.5 = 13.5 miles. He biked for 3 hours, covering 16×3 = 48 miles. His total distance was 2.5 +13.5 +48 = 64 miles, as given.

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If the suspect is found guilty, then he will go to prison.
Marrrta [24]

Answer:

If the suspect goes to prison, then he was found guilty.

Step-by-step explanation:

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valentinak56 [21]

Answer:

\frac{4x^3-3x^2+5x+6}{x+6}=4x^2-27x+167-\frac{996}{x+6}

Step-by-step explanation:

The division problem given to us is (4x^3-3x^2+5x+6)\div (x+6).

To perform the synthetic division, we write out the coefficient of the polynomial. We set the linear factor to zero and solve for x, this becomes our divisor. That is x+6=0\implies x=-6.

We carry out the synthetic division as shown in the attachment.

The result of the synthetic division is 4   -27   167   -996

The first three terms are the coefficients of the quotient and the last term is the remainder.

Therefore the quotient is q(x)=4x^2-27x+167 and the remainder is r(x)=-996.

The dividend, the quotient and the divisor are written as

\frac{4x^3-3x^2+5x+6}{x+6}=4x^2-27x+167-\frac{996}{x+6}

The correct answer is A

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Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne
Pavlova-9 [17]

Answer:

t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507  

p_v =P(t_{(15)}>1.507)=0.076

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

\bar X_{CTS}=2.35 represent the mean for the sample CTS

\bar X_{N}=1.83 represent the mean for the sample Normal

s_{CTS}=0.88 represent the sample standard deviation for the sample of CTS

s_{N}=0.54 represent the sample standard deviation for the sample of Normal

n_{CTS}=10 sample size selected for the CTS

n_{N}=7 sample size selected for the Normal

\alpha=0.01 represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis:\mu_{CTS} \leq \mu_{N}

Alternative hypothesis:\mu_{CTS} > \mu_{N}

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507  

P-value

The first step is calculate the degrees of freedom, on this case:

df=n_{CTS}+n_{N}-2=10+7-2=15

Since is a one side right tailed test the p value would be:

p_v =P(t_{(15)}>1.507)=0.076

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0
4 years ago
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