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bulgar [2K]
2 years ago
6

Robert is preheating his oven before using it to bake. The initial temperature of the oven is 65° and the temperature will incre

ase at a rate of 25° per minute after being turned on. What is the temperature of the oven 10 minutes after being turned on? What is the temperature of the oven t minutes after being turned on?
Mathematics
1 answer:
olga2289 [7]2 years ago
7 0

The temperature of the oven 10 minutes after being turned on will be 315°.

The temperature of the oven t minutes after being turned on will be 65 + 25t

<h3>How to compute the value?</h3>

The temperature of the oven 10 minutes after being turned on will be:

= 65 + (25 × t)

where t = number of minute

= 65 + (25t)

= 65 + 25(10)

= 65 + 250

= 315°

The temperature of the oven 10 minutes after being turned on will be 315°.

The temperature of the oven t minutes after being turned on will be:

= 65 + (25 × t)

= 65 + 25t

Learn more about computations on:

brainly.com/question/465834

#SPJ1

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Answer:

50.24

Step-by-step explanation:

You will do

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A = 3.14 x 2^2

A = 3.14 x 4

A = 12.56

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V = 50.24

4 0
2 years ago
A pool measuring 10 meters by 26 meters is surrounded by a path of uniform​ width, as shown in the figure. If the area of the po
igomit [66]

Based on the measurements of the pool and the path of uniform width, the area of the width and length of the path can be found to be 28.8 meters.

<h3>How to find the width and length?</h3>

First, find the area of the pool:

= Length x Width

= 10 x 26

= 260 meters ²

The area of the path can therefore be found to be:

= Total area of pool and path combined - Area of pool

= 1,092 - 260

= 832 meters²

Seeing as the path has uniform width, that means that the width is the same and the length so the width and length of the pool is:
= √area of the path

= √832

= 28.8

In conclusion, the width and length of the path are 28.8 meters.

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8 0
1 year ago
Meteorologists look for weather trends across US states in the same geographical region. Kentucky and Illinois share a border. T
olya-2409 [2.1K]

The average annual temperatures (°F) from 1991 to 2000 for Kentucky were 57.5, 55.1, 52.2, 55.6, 55.6, 54.4, 54.5, 58.3, 57.3, and 55.7.

We have to evaluate the measures of center.

Firstly, we will calculate the mean.

Mean = (Sum of all the observations)/ (Total number of observations)

Mean =  \frac{57.5+55.1+52.2+55.6+55.6+54.4+54.5+58.3+57.3+55.7}{10}

= \frac{556.2}{10}

= 55.62

Now, we will calculate mode.

Mode is a statistical term that refers to the most frequently occurring number found in a set of numbers.

Consider the data 57.5, 55.1, 52.2, 55.6, 55.6, 54.4, 54.5, 58.3, 57.3, and 55.7.

Mode = 55.6

Now, we will calculate the median.

Firstly, we will arrange in the ascending order.

52.2, 54.4, 54.4,55.1 ,55.6, 55.6, 55.7, 57.3, 57.5, 58.3

Since there are 10 terms.

Median = \frac{n}{2}th observation

Median = 5th observation.

Since, the 5th observation is 55.6.

Therefore, the median is 55.6

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4 years ago
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