The size of a television screen is given as 95 cm, correct to the nearest 5 cm.

Mathematics

1 answer:

erica [24]3 years ago

8 0

Answer:

The upper bound is 97.5 cm

Step-by-step explanation:

The upper bound is given as the value that is larger than or equal to all values in a data set, for example, in the data set, {3, 6, 16, 23, 25}, an upper bound is 25, however, where the accuracy of the data is given, the upper bound can be found by the following relation

Where the number is given to the nearest 100, add and subtract half of hundred to obtain the upper bound and lower bound respectively

For the question, given that the size of the television is given as 95 cm, correct to the nearest 5 cm, we add add half of 5 cm to get the upper bound as follows;

Upper bound = 95 cm + 5/2 cm = 97.5 cm

The upper bound = 97.5 cm.

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A school donated 283,482 cans of soup to a food kitchen if the food kitchen used 20000 cans of soup each month how many months c

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Answer:

14 months

Step-by-step explanation:

Here we are interested in calculating the number of months it will take the food kitchen to exhaust the 283,482 soup cans donated to it by a school of it uses 20,000 cans of soup.

This is pretty straightforward, trust me!

What needs to be done here is to simply divide the total number of soup cans donated by the school by the number of soup cans used by the food kitchen each month.

Mathematically, that would be;

283,482/20,000

A calculator would work here? Right?

And the division equals 14.1741

And to the nearest whole number, that is 14.

And yes there would surely be some left, carried over to the 15th month, but would that sustain them for the month? No

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The following histogram presents the amounts of silver (in parts per million) found in a sample of rocks. One rectangle from the

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Answer:

The height of the missing rectangle is 0.15

Explanation:

The image attached has the mentioned histogram.

Such histogram presents the relative frequencies for the clases [0,1], [1,2],[2,3], [4,5], and [5,6] Silver in ppm.

Only the rectangle for the class [3,4] is missing.

The height of each rectangle is the relative frequency of the corresponding class.

The relative frequencies must add 1, because each relative frequency is calculated dividing the absolute class frequency by the total number in the sample; hence, the sum of all the relative frequencies is equal to the total absolute class frequencies divided by the same number, yielding 1.

In consequence, you can sum all the known relative frequencies and subtract from 1 to get the missing relative frequency, which is the height of the missing rectangle.

1. Sum of the known relative frequencies: