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Nesterboy [21]
2 years ago
13

Round to the nearest hundredth 7.381

Mathematics
2 answers:
Soloha48 [4]2 years ago
6 0

Answer:

7.39

Step-by-step explanation:

emmainna [20.7K]2 years ago
5 0
7.39 is the answer .
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Two cylinders are similar. The radius of cylinder A is 5.6 inches. The radius of cylinder B is 1.4 inches. If the height of cyli
natita [175]
The answer is 16 inches.

Let r be a radius and h be a height of a cylinder
Since the cylinders are similar, then the ratio between their radiuses is equal to the ratio of their heights:
r1 : r2 = h1 : h2

Cylinder A:
r1 = 5.6 in
h1 = ?

Cylinder B:
r2 = 1.4 in
h2 = 4 in

5.6 : 1.4 = h1 : 4
h1 = 4 * 5.6 : 1.4
h1 = 16 inches
7 0
3 years ago
Read 2 more answers
Amber had 6 cousins. Her friend Ashley has 2 fewer than 3 times as many cousins as Amber has. Which expression represents the nu
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Answer:

Cousins Amber has: 6

(Let no. of cousins Amber has be represented as x)

Cousins Ashley has: 3x - 2

(3 X 6)- 2

18-2

16

8 0
2 years ago
Finn is walking on a treadmill at a constant pace for 30 minutes. She has programmed the treadmill for a 2-mile walk. The displa
Ymorist [56]

30*2=15miles per min. x*y=d.

4 0
3 years ago
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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
30/10 as a whole number
mrs_skeptik [129]
3.

30 divided by 10 equals 3
7 0
3 years ago
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