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MAVERICK [17]
2 years ago
13

Complete the sentence: Squares, rectangles, and rhombi are all types of ______________.

Mathematics
1 answer:
andrew-mc [135]2 years ago
5 0

Answer:

Parallelograms.

Step-by-step explanation:

Hope this helps!

If not, I am sorry.

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Find the height of the cylinder if the radius is 3 and the volume is 340
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Answer:

12.025

Step-by-step explanation:

h=v/(π*r square)

divide the volume of a cylinder by the amount of the radius squared multiplied by π(pi)

340÷(π×3square)

giving you 12.025

(original in calculator=12.02504014)

but it is 12.025

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The length of a rectangle is 11 in. longer than its width. The perimeter of the rectangle is 86 in.
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L=w+11 \\\\P=86in \\P=2L+2w \\\\ 2(w+11)+2w=86 \\\\ 2w+22+2w=86 \\\\ 4w=86-22 \\\\ 4w=64 \\\\ \boxed{w=\frac{64}{4}=16in} \\\\L=16+11 \\\\ \boxed{L=27in}
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4 years ago
The value of the pronumeral​
Sergio039 [100]

Answer: stop pay attention in class

Step-by-step explanation:

7 0
4 years ago
Andre rode his bicycle to a park located 5 1/2 miles from his house. He returned along the same route. After riding 7 1/2 miles
SSSSS [86.1K]
He needs to travel 3 1/2 more miles
7 0
3 years ago
The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are
raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

A_{i}: the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

P(A_{i})=0.1 for i = 1, 2, 3, 4, 5

The complement for A_{i} is given by

A_{i}^{$c$}: the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P(A_{i}^{$c$})=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})=(0.9)^{5}=0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P(A_{1}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5})=5×(0.1)×(0.9)^{4}=0.32805

because we have independent events.

(c) The probability that at least one contains high levels of contamination is

P(A_{1}∪A_{2}∪A_{3}∪A_{4}∪A_{5})=1-P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})=1-0.59049=0.40951

6 0
3 years ago
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