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2 years ago

How many questions would I have to miss on a 25 question test to get 88% ?

1 answer:

N76 [4]2 years ago

7 0

Let's start with the question, how many questions do you need to get 88%?

so were are looking for
?/25 = 88% and 88% is also 0.88 so we can multiply 25 by 0.88

we get 22

But this doesn't here. the question asks us how many mistakes do you need. you can just subtract 22 from 25

25-22= 3
you need 3 misses to get 88%

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10 times as many as blank tens is 30 tens

yKpoI14uk [10]

The answer to that question will be 3

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3 years ago

Quadrilateral ABCD is inscribed in this circle,

neonofarm [45]

Where’s the picture? i can’t answer it without seeing the measurements

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2 years ago

Find the perimeter. Simplify your answer.

dimaraw [331]

Answer:

30y+14

Step-by-step explanation:

add the values of all the sides together: (5y+4) + (5y+4) + (10y+3) + (10y+3) = (5y + 5y + 10y + 10y) + (4 + 4 + 3 + 3)= 30y + 14

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2 years ago

A study conducted to determine the relationship between the age, x, in years of a certain brand of motorcycle and its value, y,

kirza4 [7]

Based on the equation, y = -750·x + 8500, the estimated value of a

motorcycle that is 5 years old is <u>$4,750</u>.

<h3>How can the given equation help to estimate the value of the motorcycle?</h3>

The regression equation that models the data is; y = -750·x + 8500

Where;

y = The value of the brand of motorcycle in dollars

x = The age of the motorcycle

When the motorcycle is 5 years old, we have;

x = 5

y = -750 × 5 + 8500 = 4750

Based on the equation, the estimated value of a motorcycle that is 5 years old, y =<u> $4,750</u>

Learn more equations here:

brainly.com/question/20475360

4 0

2 years ago

Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago