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dimaraw [331]
1 year ago
12

What is the greatest possible percent error in calculating the volume of a box measured to the nearest inch with sides of 5 in.,

10 in., and 10 in.? Round to the nearest percent.
Mathematics
2 answers:
Andre45 [30]1 year ago
8 0
If each of the side lengths was one inch off, the measurement of the box would be 6*11*11 which would equal 726 inches instead of 500. 500/726≈0.69, so the answer to the nearest percent is 31%. Hope this helps!
Mashutka [201]1 year ago
7 0
The answer is 20 percent
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Kite A B C D is shown. Lines are drawn from point A to point C and from point B to point D and intersect. In the kite, AC = 10 a
Oduvanchick [21]

Answer:

30 u^{2}

Step-by-step explanation:

The computation of the area of kite ABCD is shown below:

Given data

AC = 10 ;

BD = 6

As we can see from the attached figure that the Kite is a quadrilateral as it involves two adjacent sides i.e to be equal

Now the area of quadrilateral when the diagonals are given

So, it is

\text { area of kite }=\frac{1}{2} \times d_{1} d_{2}

where,

d_{1}=10\ and\ d_{2}=6

So, the area of the quadrilateral is

=\frac{1}{2}(10)(6)\\\\=30 u^{2}

4 0
1 year ago
The NEC permits 20% of the cross sectional area in a wireway to be occupied by conductors.for an 8"×8"×6' long wireway the condu
Zanzabum

Answer:

Step-by-step explanation:

Sjaoa

8 0
1 year ago
In a test of a printed circuit board using a random test pattern, an array of 16 bits is equally likely to be 0 or 1. Assume the
Doss [256]

Answer:

A) b) and c) answer has been explained below.

Step-by-step explanation:

For the question would use 1's and 0's and taking probabilities to be equal for both using random test pattern whose formula is when p = q

then Simplify to

P[k] =  nCk /2^n

A.Probability that all bits are 1s

16c16/2^16 = 1/65536

B. Probability that all bits are 0s

16c0/2^16 = 1/65536

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8 0
2 years ago
Find the first three iterates of the function f(z) = 2z + (3 - 2i) with an initial value of
nikklg [1K]

Answer:

a.  5 + 2i, 13 + 2i, 29 + 2i

Step-by-step explanation:

We'll use the formula  f(z) = 2z + (3 - 2i)  for each iteration. The output of the first iteration will be come the input of the second iteration, and so on.

So, we start with z0 = 1 + 2i and we plug that into the base equation:

z0 = 1 + 2i ==> f(z) = 2(1 + 2i) + 3 - 2i = 2 + 4i + 3 - 2i = 5 + 2i

z1 = 5 + 2i ==> f(z) = 2(5 + 2i) + 3 - 2i = 10 + 4i + 3 - 2i = 13 + 2i

z2 = 13 + 2i ==> f(z) = 2(13 + 2i) + 3 - 2i = 26 + 4i  + 3 - 2i = 29 + 2i

z3 = 29 + 2i

8 0
1 year ago
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An adult meerkat weighed 0.776 kilograms . this is 7.47 hectograms more than the weight of a baby meerkat . how much dose a baby
Firlakuza [10]
Let x = weight of adult meerkat, y=weight of baby meerkat, Find y.
x=0.776 kg
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Since 1 hectogram = 0.1 kg,
Therefore the weight of baby meerkat is, y= 0.029 kg
7 0
2 years ago
Read 2 more answers
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