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3 years ago

148/27 as a mixed number

Mathematics

1 answer:

Gnoma [55]3 years ago

8 0

$\frac{148}{27} = 5 \frac{13}{27}$
......................................

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In Ms.Khans class,22 out of 25 students hand in their projects on time. What percent of the class hand in their projects on time

Nikitich [7]

88 % students of the class hand in their projects on time

<em><u>Solution:</u></em>

Given that, Ms.Khans class, 22 out of 25 students hand in their projects on time

Total number students = 25

Students hand in their projects on time = 22

<em><u>To find: percent of the class hand in their projects on time</u></em>

$\text{ percent of class hand in time } = \frac{\text{Students hand in projects on time}}{\text{total students}} \times 100$

Substituting the values, we get

$\text{ percent of class hand in time } = \frac{22}{25} \times 100\\\\\text{ percent of class hand in time } = 0.88 \times 100\\\\\text{ percent of class hand in time } = 88$

Thus 88 % students of the class hand in their projects on time

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3 years ago

Solve X^2 - 3x - 10 = 0 by using the quadratic formula

stiv31 [10]

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Answer:

Step-by-step explanation:

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4 years ago

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Can anyone help me with this I need a answer for this before 11:30am?

algol [13]

Hello from MrBillDoesMath!

Answer:

x = 2

Discussion:

See attachment for full detail.

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4 years ago

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PtichkaEL [24]

4398200 because get rid of the ones after 200

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4 years ago

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In a certain lake, trout average 12 in. in length with standard deviation 2.75 in. and the bass average 4 lb. in weight with sta

NISA [10]

Answer:

The bass fish was the better catch

Step-by-step explanation:

From the question we are told that

The population mean for trout is $\mu_1 = 12 \ in$

The standard deviation is $\sigma_1 = 2.75 \ in$

The population mean for base is $\mu _2 = 4 \ lb$

The standard deviation is $\sigma_2 = 0.8 \ lb$

The number of trout caught $x_1 = 18$

The number of bass caught $x_2 = 6$

Generally z-value(standardized value ) for the of number trout caught is mathematically represented as

$z_1 = \frac{x_1 - \mu_1}{\sigma_1 }$

substituting value

$z_1 = \frac{18 - 12}{2.75 }$

$z_1 = 2.18$

Generally z-value(standardized value ) for the of number bass caught is mathematically represented as

$z_2 = \frac{x_2 - \mu_2}{\sigma_2 }$

substituting value

$z_2 = \frac{6 - 4}{0.8 }$

$z_2 = 2.5$

From our calculation we see that $z_2 > z_1$

The fish that was the better catch is the bass fish

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3 years ago