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FinnZ [79.3K]
3 years ago
5

Twenty different statistics students are randomly selected. for each of​ them, their body temperature ​( degrees °​c) is measure

d and their head circumference​ (cm) is measured. if it is found that r=​0, does that indicate that there is no association between these two​ variables?
Mathematics
1 answer:
Trava [24]3 years ago
8 0
Yes, it does.

An r-value, or correlation coefficient, goes from -1 to positive 1.  -1 indicates a perfect decreasing match and 1 indicates a perfect increasing match.  0 indicates no relationship between the variables.
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On a separate piece of paper, draw the different ways you can divide the given shape into simpler shapes. Describe each result.
marishachu [46]

Answer:

there will be 4 cornered shapes 1 rectangle that is almost a square and the shape I Forgot TvT

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3 years ago
16 feet perimeter 5 feet long what is the width
miskamm [114]
P=2L+2W
L=Length
W=Width
P=Perimeter=16
So you have:
16=2×5+2w
Multiply 2 and 5 and get:
16=10+2w
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The width is 3.


8 0
3 years ago
Plz solve this question <br>Its urgent.....​
Maksim231197 [3]

Answer:

Step-by-step explanation:

tan Θ + tan 2Θ + √3 tan Θ tan 2Θ = √3

tan Θ + tan 2Θ = √3 - √3 tan Θ tan 2Θ

tan Θ + tan 2Θ = √3 ( 1 - tan Θ tan 2Θ)

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tan(Θ  + 2Θ) = √3

tan 3Θ = tan (\frac{\pi }{3})           we know tan Θ = tan α; Θ = nΠ + α, n belongs to z

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4 0
3 years ago
200 + 200m - 100 - 100m + 100
slava [35]

Answer:

200+100m

Step-by-step explanation:

Simplified by collecting like terms

5 0
2 years ago
“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and wor
Artist 52 [7]

Answer:

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

C=2\pi r

<u><em>Find the radius of the sphere with a 5.50-meter circumference</em></u>

For C=5.50\ m

assume

\pi =3.14

substitute and solve for r

5.50=2(3.14)r

r=5.50/[2(3.14)]=0.88\ m

<u><em>Find the radius of the sphere with a 7.85-meter circumference</em></u>

For C=7.85\ m

assume

\pi =3.14

substitute and solve for r

7.85=2(3.14)r

r=7.85/[2(3.14)]=1.25\ m

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

SA=4\pi r^{2}

<u><em>Find the surface area of sphere with a 5.50-meter circumference</em></u>

For r=0.88\ m

assume

\pi =3.14

substitute

SA=4(3.14)(0.88)^{2}

SA=9.73\ m^{2}

<u><em>Find the surface area of sphere with a 7.85-meter circumference</em></u>

For r=1.25\ m

assume

\pi =3.14

substitute

SA=4(3.14)(1.25)^{2}

SA=19.63\ m^{2}

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

<u><em>Find the cost of sphere with a 5.50-meter circumference</em></u>

9.73*(92)=\$895.16

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

<u><em>Find the cost of sphere with a 7.85-meter circumference</em></u>

19.63*(92)=\$1,805.96

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

6 0
3 years ago
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