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Yuliya22 [10]
3 years ago
9

Which of the following values have 3 significant figures? Check all that apply.

Mathematics
2 answers:
iVinArrow [24]3 years ago
7 0
Option c) and d) have 3 significant figures
balu736 [363]3 years ago
3 0

OPTION C AND D

99.8 AND 458 are the values have 3 significant figures

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What is 5:1 as a fraction?
vodomira [7]
The fraction would be 5/1.
7 0
3 years ago
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The coordinates of rhombus ABCD are A(–4, –2), B(–2, 6), C(6, 8), and D(4, 0). What is the area of the rhombus? Round to the nea
a_sh-v [17]
Check the picture below.

so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad 
C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2}
\\\\\\
AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\
-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad 
D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2}
\\\\\\
BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2}
\\\\\\
BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}

that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.

namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,

\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right]
\\\\\\
4[15]\implies 60

7 0
3 years ago
Read 2 more answers
3 3/5 + 9 3/5 answer with a mixed number in simplest form
Romashka [77]

Both fractions have the same denominator so it's going to be easy. First set up your equation:

3 \frac{3}{5} + 9 \frac{3}{5}

Add the whole numbers:

3 + 9 = 12

Now add the fractions:

\frac{3}{5} + \frac{3}{5} = \frac{6}{5}

add both together:

12 + \frac{6}{5} = 12 \frac{6}{5}

\frac{6}{5} is an improper fraction so change it:

12 \frac{6}{5} = 13 \frac{1}{5}

Since 6 is one more than 5, add 1 to the whole number and subtract the numerator and denominator(6 - 5 = 1) and make the remaining the new numerator. That leaves you with 13 \frac{1}{5}

Your answer is 13 \frac{1}{5}

3 0
3 years ago
Scores for a standardized test are normally distributed with a mean of 350 and a standard deviation of 25. If 1793 students scor
Sergeu [11.5K]

Answer:

I think this is right 50 but I don't know I am 3rd grade

Step-by-step explanation:

<h2>so thaso look 350 so minus that is 50s my answer</h2>
6 0
2 years ago
[ Easy Math ] Help me pls
goldfiish [28.3K]

Answer:

1) you're going to have to flip the coins (or fake numbers) for the experimental trials.

2) for the theoretical, there is 1/2 chance for heads or tails with each toss, so you'd expect that out of 10 tosses, 5 heads, 5 tails. out of 100 tosses- 50 heads, 50 tails.

When tossing 2 coins- 1/2×1/2 = 1/4 (25%) chance that 2 heads, 2 tails, or 1 heads & 1 tails. Deviation value comes from after you done your flipping and recorded your data. So if on 100 flips you actually got 50 and 50 (rarely us that exact ;), the deviation from the expected of 50/50 would be 0.00. If however you flipped 100 heads or 100 tails (impossible), then the deviation value would be 1.00.

|(100-50)| ÷ 50 = 50÷50 = 1.00

So usually you may have data like: 47/53 or something a little off than 50/50, making deviation |(47-50)| ÷ 50 = 3÷50 = 0.06.

Now the number of flips is important for the outcome! So if a coin toss if 10 times had 4 heads, 6 tails, the deviation value would be:

|(4-5)| ÷ 5 = 1÷5 = 0.20

So increasing the # flips DECREASES the deviation value!!

Whether it's from 10 to 100, or from 100 to 200. Look at my example of how the 10-flip deviation of 0.20 decreased to 0.06 with 100-flip

7 0
3 years ago
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