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bearhunter [10]
3 years ago
12

To rent a certain meeting room, a college charges a reservation fee of $41 and an additional fee of $7.90 per hour. The chemistr

y club wants to spend at most
$96.30 on renting the meeting troom.
What are the possible amounts of time for which they could rent the meeting room?
Use t for the number of hours the meeting room is rented, and solve your inequality for 1.

Mathematics
1 answer:
Lemur [1.5K]3 years ago
4 0

Answer:

41

Step-by-step explanation:

Rounding

Your cost just for the room is 41

You pay 7.90 for Your hours of being there

And your Budget is 96.30

the equation wants you for figure how many hours would you be there at the establishment that you would pay your full budget 96.30 or closer and You would only pay 41.00

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almond37 [142]

Answer:

Step-by-step explanation:

32 is the correct answer

6 0
2 years ago
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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
2 years ago
Differential cos^2x dy/dx =e^y-tanx​
gulaghasi [49]

Answer:

y=t−1+ce

−t

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Given, cos

2

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⇒

dx

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2

x=tanxsec

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x ....(1)

Here P=sec

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Multiplying (1) by I.F. we get

e

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dx

dy

+e

tanx

ysec

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x=e

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ye

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Put tanx=t⇒sec

2

xdx=dt

∴ye

t

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t

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mario62 [17]

Answer:

400


Step-by-step explanation:


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Answer:

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