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Karolina [17]
3 years ago
10

A set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. Find the percent of data between 4.2

and 5.1.
A)about 68%
B)about 34%
C)about 16%
D)about 13.5%
Mathematics
1 answer:
Yuliya22 [10]3 years ago
7 0

A set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. Find the percent of data between 4.2 and 5.1.

Answer: The correct option is B) about 34%

Proof:

We have to find P(4.2

To find P(4.2, we need to use z score formula:

When x = 4.2, we have:

z = \frac{x-\mu}{\sigma}

          =\frac{4.2-5.1}{0.9}=\frac{-0.9}{0.9}=-1

When x = 5.1, we have:

z = \frac{x-\mu}{\sigma}

          =\frac{5.1-5.1}{0.9}=0

Therefore, we have to find P(-1

Using the standard normal table, we have:

P(-1= P(z

                               =0.50-0.1587

                               =0.3413 or 34.13%

                               = 34% approximately

Therefore, the percent of data between 4.2 and 5.1 is about 34%

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Step-by-step explanation:

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The radius of a cylinder is 7 c m and its height is 10 c m. Find curved surface area and volume.​
erma4kov [3.2K]

Answer:

  • CSA of the cylinder = 440 sq. cm

  • Volume of the cylinder = 1540 cu. cm

\\

Step-by-step explanation:

Given:

  • Radius of the cylinder = 7 cm
  • Height of the cylinder = 10 cm

\\

To Find:

  • Curved surface area
  • Volume

\\

Solution:

\\

Using formula:

\dashrightarrow \:  \:  { \underline{ \boxed{ \pmb{ \sf{ \purple{CSA  \: of  \: cylinder = 2\pi rh}}}}}}  \:  \star \\  \\

<em>Substituting the required values: </em>

\\

\dashrightarrow \:  \:  \sf CSA {(cylinder)} = 2 \times \dfrac{22}{7} \times 7 \times  10 \\  \\  \\  \dashrightarrow \:  \:  \sf CSA {(cylinder)}  = 2 \times 22 \times 10 \\  \\ \\   \dashrightarrow \:  \:  \sf CSA {(cylinder)} = 44 \times 10 \\  \\  \\  \dashrightarrow \:  \:  \sf CSA {(cylinder)}  = 440 \:  {cm}^{2}  \\ \\ \\

Now,

\dashrightarrow \:  \:  { \underline{ \boxed{ \pmb{ \sf{ \purple{Volume {(cylinder)}=  \pi {r}^{2} h}}}}}}  \:  \star \\  \\ \\

<em>Substituting the required values, </em>

\\

\dashrightarrow \:  \:   \sf Volume {(cylinder)}=  \dfrac{22}{7}  \times  {(7)}^{2}  \times 10 \\ \\  \\  \dashrightarrow \:  \:   \sf Volume {(cylinder)}=  \frac{22}{7}  \times 49  \times 10 \\ \\ \\  \dashrightarrow \:  \:   \sf Volume {(cylinder)}= 22 \times 7 \times 10 \\ \\  \\  \dashrightarrow \:  \:   \sf Volume {(cylinder)}= 1540 {cm}^{3}  \\ \\ \\

Hence,

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  • Volume of the cylinder = 1540 cu. cm
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