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Arada [10]
3 years ago
7

the average child will wear down 738 crayons by his or her 10th birthday find the number of boxes of 64 crayons this is equivale

nt to round to the nearest tenth
Mathematics
2 answers:
kupik [55]3 years ago
8 0

\frac{738}{64}  = 11.531
around 11.5 boxes of 64 crayons
Rus_ich [418]3 years ago
7 0
11.5 boxes. 738/64 so your answer would be 11.53, but you said nearest 10th so 11.5
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Compare and contrast the perpendicular bisectors and angle bisectors of a triangle. How are they alike? How are they different?
AleksandrR [38]

<span>1 perpendicular bisector divides 1 side of the triangle into two equal lengths at a 90 degree angle. So if there's is three perpendicular bisectors, then it splits all three sides the triangles into equal lengths.

Angle bisectors splits an angle of a triangle into two even degrees.  
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<span>EX: if it is an equilateral triangle, each angle will be sixty degrees. Therefore, ONE angle bisector will split ONE of the angles into even degrees of thirty degrees. </span>

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3 years ago
A rectangle is 1/3 yd long and 2/3 yd wide
ICE Princess25 [194]
It might be 2/9 sorry if it wasn’t
3 0
3 years ago
Someone i need help i cant expand these expressions
vovangra [49]

Answer: The answer is 5x+6

Step-by-step explanation:

1/3(9x+16+2)+2x

= 1/3(9x)+1/3(16)+1/3(2)+2x

=3x/1+16/3+2/3+2x/1

=3x/1+2x/1+16/3+2/3

Take the L.C.M

=3x+2x/1+16+2/3

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4 0
3 years ago
I need an accurate answer please
kotykmax [81]

x = –6

Solution:

Given expression is \sqrt[3]{x-2}+5=3.

Step 1: Isolate the radical by subtracting 5 from both sides of the equation.

\Rightarrow\sqrt[3]{x-2}+5-5=3-5

\Rightarrow\sqrt[3]{x-2}=-2

Step 2: Cube both sides of the equation to remove the cube root.

\Rightarrow(\sqrt[3]{x-2})^3=(-2)^3

Cube and cube root get canceled in left side of the equation.

\Rightarrow\ x-2=-8

Step 3: To solve for x.

Add 2 on both sides of the equation.

\Rightarrow\ x-2+2=-8+2

\Rightarrow\ x=-6

Hence the solution is x = –6.

6 0
3 years ago
Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh
s344n2d4d5 [400]

Answer: The point estimate for the mean weight of all spawning Chinook salmon in the Columbia River is the mean weight from your sample (\overline{x}) = 31.2 pounds .

Step-by-step explanation:

The weights of spawning Chinook salmon in the Columbia river are normally distributed.

Given :  The mean weight from your sample is 31.2 pounds with a standard deviation of 4.4 pounds.

The point estimate for the mean (\mu) weight of all spawning Chinook salmon in the Columbia River is the mean weight from your sample (\overline{x}) i.e. is 31.2 pounds because the best point estimate for population mean is the sample mean. [in normal distribution]

8 0
3 years ago
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