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Setler [38]
3 years ago
12

What is the area and circumference of a circle with a radius of 6.5 centimeters​

Mathematics
2 answers:
Ierofanga [76]3 years ago
8 0

Answer:

Area: 132.73cm².

Circumference: 40.84cm².

Step-by-step explanation: 6.5 centimeter radius is equal to 132.73cm area, and 40.84cm circumference.

I hope this helps!

allochka39001 [22]3 years ago
4 0

Answer:

Step-by-step explanation:

Area : about 132.73

Circumference : about 40.84

Hope this helped!

Please mark me brainliest

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Which equation matches the given points? (0, 7.6), (1, 6.2), (2, 4.8), (3, 3.4), (4,2)
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The equation that matches the given points is g(x) = -1.4x + 7.6

The standard form of a linear  equation is expressed as g(x)=mx + b

  • m is the slope of the line
  • b is the y-intercept:

Using the coordinate points (3, 3.4), (4,2)

m=\frac{2-3.4}{4-3}\\m=\frac{-1.4}{1}\\m=-1.4

Substitute m = -1.4 and the coordinate (4, 2) into the formula:

2=4(-1.4)+b\\2=-5.6+b\\b=2+5.6\\b=7.6

Get the required equation:

g(x)=mx+b\\g(x)=-1.4x+7.6

Hence the equation that matches the given points is g(x) = -1.4x + 7.6

Learn more on equation of a line here; brainly.com/question/9351428

4 0
2 years ago
Match the description with the ratio. 1. There are three silver bows for every four red bows in a bag. 3:10 2. There are four re
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The composite shape is made up of two different rectangles and two different triangles. What is the area of the composite shape
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Answer:

43 unit²

Step-by-step explanation:

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Consider an urn containing 8 white balls, 7 red balls and 5 black balls.
weqwewe [10]

Answer + Step-by-step explanation:

1) The probability of getting 2 white balls is equal to:

=\frac{8}{20} \times \frac{7}{19}\\\\= 0.147368421053

2) the probability of getting 2 white balls is equal to:

=C^{2}_{5}\times (\frac{8}{20} \times \frac{7}{19}) \times (\frac{12}{18} \times \frac{11}{17} \times \frac{10}{16})\\=0.397316821465

3) The probability of getting at least 72 white balls is:

=C^{72}_{150}\times \left( \frac{8}{20} \right)^{72}  \times \left( \frac{7}{20} \right)^{78}  +C^{73}_{150}\times \left( \frac{8}{20} \right)^{73}  \times \left( \frac{7}{20} \right)^{77}  + \cdots +C^{149}_{150}\times \left( \frac{8}{20} \right)^{149}  \times \left( \frac{7}{20} \right)^{1}  +\left( \frac{8}{20} \right)^{150}

=\sum^{150}_{k=72} [C^{k}_{150}\times  \left( \frac{8}{15} \right)^{k}  \times \left( \frac{7}{15} \right)^{150-k}]

5 0
1 year ago
Read 2 more answers
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