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kakasveta [241]
3 years ago
15

A hot air balloon rose from a height of 100 m to 400 m in 3 minutes. What was the balloon’s rate of change?

Mathematics
1 answer:
maria [59]3 years ago
7 0

Answer:

1.666666667 meter per sec^2

Step-by-step explanation:

I feel like this is physic

using acceleration formla Vf-Vi/t

so 400-100/180sec

300/180

1.666666667 meter per sec^2

Hey  im still learning tell me if you feel like im wrong OwO

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41/50 is closer to 9/11 or 10/11 and explain
Stolb23 [73]
<span>Given number is 41/50,
Now, let’s find which one is closer to the given choices:
=> 9/11
=> 10/11
Notice that the given number and given choices are fraction numbers. In order know which one is closer to the given number, we need to convert these numbers into a decimal.
Note, that in converting a fraction number to a decimal is we need to divide our numerator by our denominator to get the answer.
=> 41/50
=> 0.82
The value of 41/50 is 0.82. Now, let’s find the value of the given choices.
=> 9/11
=> 0.818 or 0.82

=> 10/11
=>0.909 or 0.91

Thus, the correct answer is 9/11 is equals to 41/50.</span>



3 0
3 years ago
The club charges $10 per hour for snowboard lessons and $40 for the lift ticket and a snowboard rental.
S_A_V [24]
Is there more to this question?
5 0
3 years ago
I need this answered in ONE minute
Helen [10]

Answer:

x^4 + 4x^3 + 6x^2 + 7x + 2

Step-by-step explanation:

We are asked to multiply the given polynomials.

(x^ 2 + 3x + 1) \times (x^2 + x + 2)

Multiply each term of the first polynomial to each term of the second polynomial.

x^ 2  \times (x^2 + x + 2) = x^4 + x^3 + 2x^2

3x  \times (x^2 + x + 2) = 3x^3 + 3x^2 + 6x

1 \times (x^2 + x + 2) =  x^2 + x + 2

Add the results

(x^4 + x^3 + 2x^2) + (3x^3 + 3x^2 + 6x) + ( x^2 + x + 2)

Combine the like terms

x^4 + 4x^3 + 6x^2 + 7x + 2

The answer is written in descending powers of x.

7 0
3 years ago
Suppose a company produces report indicates that the surveyed distance between two points is 1200 feet with a margin of error of
Otrada [13]
<h2>Hello!</h2>

The answers are:

MaximumDistance=1200.1ft\\MinimumDistance=1199.9ft

<h2>Why?</h2>

Since we are given the margin of error and it's equal to ±0.1 feet, and we know the surveyed distance, we can calculate the maximum and minimum distance. We must remember that margin of errors usually involves and maximum and minimum margin of a measure, and it means that the real measure will not be greater or less than the values located at the margins.

We know that the surveyed distance is 1200 feet with a margin of error of ±0.1 feet, so, we can calculate the maximum and minimum distances that the reader could assume in the following way:

MaximumDistance=ActualDistance+0.1feet\\\\MaximumDistance=1200feet+0.1feet=1200.1feet

MinimumDistance=ActualDistance-0.1feet\\\\MaximumDistance=1200feet-0.1feet=1199.9feet

Have a nice day!

4 0
3 years ago
The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room
Alexandra [31]

Answer:

\bar x = 260.1615

\sigma = 70.69

The confidence interval of standard deviation is: 53.76 to 103.25

Step-by-step explanation:

Given

n =20

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

\bar x = \frac{\sum x}{n}

So, we have:

\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}

\bar x = \frac{5203.23}{20}

\bar x = 260.1615

\bar x = 260.16

Solving (b): The standard deviation

This is calculated as:

\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}

\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}\sigma = \sqrt{\frac{94938.80}{19}}

\sigma = \sqrt{4996.78}

\sigma = 70.69 --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

c =0.95

So:

\alpha = 1 -c

\alpha = 1 -0.95

\alpha = 0.05

Calculate the degree of freedom (df)

df = n -1

df = 20 -1

df = 19

Determine the critical value at row df = 19 and columns \frac{\alpha}{2} and 1 -\frac{\alpha}{2}

So, we have:

X^2_{0.025} = 32.852 ---- at \frac{\alpha}{2}

X^2_{0.975} = 8.907 --- at 1 -\frac{\alpha}{2}

So, the confidence interval of the standard deviation is:

\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} } to \sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }

70.69 * \sqrt{\frac{20 - 1}{32.852} to 70.69 * \sqrt{\frac{20 - 1}{8.907}

70.69 * \sqrt{\frac{19}{32.852} to 70.69 * \sqrt{\frac{19}{8.907}

53.76 to 103.25

8 0
3 years ago
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