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xeze [42]
1 year ago
15

Find the area of the trapezoidal cross-section of the irrigation canal shown below. Your answer will be in terms of h, w, and θ.

Mathematics
1 answer:
Setler [38]1 year ago
5 0
The answer is :h^2/tanθ+wh
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What are the solution(s) to the quadratic equation 50 – x2 = 0? x = ±2Plus or minus 2 StartRoot 5 EndRoot x = ±6Plus or minus 6
Serga [27]

Question:

What are the solution(s) to the quadratic equation 50 – x2 = 0?

A) x = ±2Plus or minus 2 StartRoot 5 EndRoot

B) x = ±6Plus or minus 6 StartRoot 3 EndRoot

C) x = ±5Plus or minus 5 StartRoot 2 EndRoot

D) no real solution

Answer:

C) x = ±5Plus or minus 5 StartRoot 2 EndRoot

3 0
2 years ago
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Jing spent 1/3 of her money on a pack of pens, 1/2 of her money on a pack of markers, and 1/8 of her money on a pack of pencils.
Reika [66]

As per the problem

Jing spent \frac{1}{3} of her money on a pack of pens.

\frac{1}{2} of her money on a pack of markers.

and \frac{1}{8} of her money on a pack of pencils.

Total fraction of money spent cab be given as below

Fraction of Money Spent =\frac{1}{3} +\frac{1}{2}+\frac{1}{8}

Take the LCD of denominator, we get LCD of (3,2,8)=24

Fraction of Money Spent =\frac{8+12+3}{24} =\frac{23}{24} \\\\

\\ \text{Hence fraction of Money Spent }=\frac{23}{24} \\ \\ \text{Fraction of Money left}=1-\frac{23}{24} \\ \\ \text{Simplify, we get}\\ \\ \text{Fraction of Money left}=\frac{24-23}{24} \\  \\ \text{Fraction of Money left}=\frac{1}{24}

3 0
2 years ago
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Jala put $600 in an interest bearing account with a annual compound interest rate of 5%. Jala determined that after seven years,
Inessa [10]

Answer:

7.4 years would be the correct answer

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2 years ago
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According to a "how to stop bullying" Web site, 15% of students report experiencing bullying one to three times within the most
AlekseyPX

Answer:

His 95% confidence interval is (0.065, 0.155).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 186, \pi = 0.11

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.11 - 1.96\sqrt{\frac{0.11*0.89}{186}} = 0.065

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.11 + 1.96\sqrt{\frac{0.11*0.89}{186}} = 0.155

His 95% confidence interval is (0.065, 0.155).

4 0
1 year ago
The lateral surface area is _ square inches
lidiya [134]

Answer:

180

Step-by-step explanation:

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