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ycow [4]
3 years ago
11

During your cycling trip, you and your friends often stayed in hostels. The prices of these are listed below.

Mathematics
1 answer:
kiruha [24]3 years ago
7 0

Answer:

  • a) NTRD standard price = $26.30
  • b) range: $11.20 to $33.00, a range of $21.80

Step-by-step explanation:

a) The average is the sum divided by the number of contributors. Let n represent the standard price at Night-Time Rest Days. Then ...

... ($22.45 +18.55 +24.50 +11.20 +n)/5 = $20.60

... 76.70 + n = 103.00 . . . . multiply by 5 and collect terms

... n = $26.30 . . . . standard price at NTRD

b) The lowest standard price is $11.20. The highest breakfast price is $33.00. The range is the difference between these ...

... $33.00 -11.20 = $21.80 . . . . range in cost for a single night

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Step-by-step explanation:

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as a whole number = 5

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Find the next three terms of the sequence: 10, 6, 2, .
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Step-by-step explanation:

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3 years ago
Which of the following Inequalities is graphed on the coordinate plane?
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Determine the slope between the points (-1, -5) and (5, 4)
Basile [38]

Answer:

Slope = 9/6

Step-by-step explanation:

Step 1: We have formula to find the slope if we are given two point.

<h3>slope = \frac{(y2 - y1)}{(x2 - x1)}</h3>

Given points (-1, -5) and (5, 4)

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Thank you.

Hope you will understand this.

4 0
3 years ago
A cake is removed from a 310°F oven and placed on a cooling rack in a 72°F room. After 30 minutes the cake's temperature is 220°
Fynjy0 [20]

Answer:

The time is 135 min.

Step-by-step explanation:

For this situation we are going to use Newton's Law of Cooling.

Newton’s Law of Cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding medium and is given by

T(t)=C+(T_0-C)e^{kt}

where,

C = surrounding temp

T(t) = temp at any given time

t = time

T_0 = initial temp of the heated object

k = constant

From the information given we know that:

  • Initial temp of the cake is 310 °F.
  • The surrounding temp is 72 °F.
  • After 30 minutes the cake's temperature is 220 °F.

We want to find the time, in minutes, since the cake's removal from the oven, at which its temperature will be 100°F.

To do this, first, we need to find the value of k.

Using the information given,

220=72+(310-72)e^{k\cdot 30}\\\\72+238e^{k30}=220\\\\238e^{k30}=148\\\\e^{k30}=\frac{74}{119}\\\\\ln \left(e^{k\cdot \:30}\right)=\ln \left(\frac{74}{119}\right)\\\\k\cdot \:30=\ln \left(\frac{74}{119}\right)\\\\k=\frac{\ln \left(\frac{74}{119}\right)}{30}

T(t)=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}

Next, we find the time at which the cake's temperature will be 100°F.

100=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}\\72+238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=100\\238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=28\\e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=\frac{2}{17}\\\ln \left(e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}\right)=\ln \left(\frac{2}{17}\right)\\\frac{\ln \left(\frac{74}{119}\right)}{30}t=\ln \left(\frac{2}{17}\right)\\t=\frac{30\ln \left(\frac{2}{17}\right)}{\ln \left(\frac{74}{119}\right)}\approx 135.1

4 0
3 years ago
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