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Ivenika [448]
3 years ago
12

Two cylinders are similar. The surface area of one is 49 cm ^ 2 , and the surface area of the other is 121 cm ^ 2 . Find the sca

le factor between them.

Mathematics
2 answers:
dalvyx [7]3 years ago
8 0

Answer:

7 : 17.29

Step-by-step explanation:

==>Given:

Two similar cylinders:

Surface area of Cylinder A = 49cm²

Surface area of Cylinder B = 121cm²

==>Required:

Scale factor between both cylinders

==>Solution:

We can easily get the scale factor between both by taking the following steps:

=>Set up a ratio of their surface area:

Smaller cone surface area : larger cone surface area

49:121

= 49/121

=>Simplify the ratio gotten:

Dividing the denominator and the numerator by 7, would give us,

7/17.29

= 7:17.29

Sloan [31]3 years ago
6 0

Answer:11

Step-by-step explanation:

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In a certain Algebra 2 class of 30 students, 19 of them play basketball and 12 of them play baseball. There are 8 students who p
Alenkinab [10]

Answer:

Probability that a student chosen randomly from the class plays basketball or baseball is  \frac{23}{30} or 0.76

Step-by-step explanation:

Given:

Total number of students in the class = 30

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Number of students who plays base ball = 12

Number of students who plays base both the games = 8

To find:

Probability that a student chosen randomly from the class plays basketball or baseball=?

Solution:

P(A \cup B)=P(A)+P(B)-P(A \cap B)---------------(1)

where

P(A) = Probability of choosing  a student playing basket ball

P(B) =  Probability of choosing  a student playing base ball

P(A \cap B) =  Probability of choosing  a student playing both the games

<u>Finding  P(A)</u>

P(A) = \frac{\text { Number of students playing basket ball }}{\text{Total number of students}}

P(A) = \frac{19}{30}--------------------------(2)

<u>Finding  P(B)</u>

P(B) = \frac{\text { Number of students playing baseball }}{\text{Total number of students}}

P(B) = \frac{12}{30}---------------------------(3)

<u>Finding P(A \cap B)</u>

P(A) = \frac{\text { Number of students playing both games }}{\text{Total number of students}}

P(A) = \frac{8}{30}-----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}

P(A \cup B)= \frac{31}{30} -\frac{8}{30}

P(A \cup B)= \frac{23}{30}

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