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viktelen [127]
3 years ago
12

Len estimated the sum of 16.23, 215.7, and 110.3 by rounding each number to the nearest whole.

Mathematics
1 answer:
kozerog [31]3 years ago
6 0

Estimated sum = 342

Actual sum = 342.23

Solution:

<u>Rule to rounding decimal number to whole:</u>

If the number after the decimal point is greater than or equal to 5, then add one number to the whole number and remove the decimal terms.

If the number after the decimal point is less than 5, then remove the decimal term and the whole number is as it is the same.

Nearest whole of 16.23 = 16 (decimal is less than 5, so as it is same whole)

Nearest whole of 215.7 = 216 (decimal is greater than 5, so add 1 number)

Nearest whole of 110.3 = 110 (decimal is less than 5, so as it is same whole)

Estimated sum = 16 + 216 + 110

Estimated sum = 342

Actual sum = 16.23 + 215.7 + 110.3

Actual sum = 342.23

Hence Len's estimated sum of the numbers is 342 and the actual sum of the numbers is 342.23.

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Kiran scored 223 fewer points in a computer game than Tyler. Kiran scored 409 points. Which equation would Kiran use to determin
LenKa [72]

Answer:

The equation is:

409 = T - 223

And the solution is:

T = 632

Step-by-step explanation:

Given

Represent Kiran with K and Tyler with T

Kiran score is represented as:

K = 409

Also, Kiran scored 223 less than Tyler.

This is represented as:

K = T - 223

Substitute 409 for K

409 = T - 223

Make T the subject

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4. Line t and AECA and AFDB are shown on the coordinate plane
julia-pushkina [17]

Answer:

A, B, C, E

Step-by-step explanation:

It can be seen from the figure that the points A, B, C and D, all are lying in the line t.

=> So that it can be concluded that AC and BC and BD have the slopes which are equal to each other and also equal to the slop of line t

So that all answer A, B, C are true.

In addition, as FD is parallel with x - axis, so that slope of the line t is equal to <em>tan angel FDB </em>

As FDB is the right triangle with BFD = 90°

=> tan angel FDB = FB/ FD (tan of an acute angel in the right triangle = opposite side/ adjacent side)

=> Slope of the line t is equal to FB/ FD

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The addison see to the horizon at 2 root 2mi.

We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.

We have to find the how much farther can addison see to the horizon

<h3>Which equation we get from the given condition?</h3>

d=\sqrt{\frac{3h}{2} }

Where, we have

d- the distance they can see in thousands

h- their eye-level height in feet

For Kaylib

d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)

For Addison h=85(1/3)

d=\sqrt{\frac{3\times 85\frac{1}{3} }{2} }\\d\sqrt{\frac{256}{2} } \\d=\sqrt{128} \\d=8\sqrt{2} .....(2)

Subtracting both distances we get

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Therefore, the addison see to the horizon at 2 root 2mi.

To learn more about the eye level visit:

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