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3 years ago

I need to know a equivalent fraction for number 7

Mathematics

1 answer:

AlekseyPX3 years ago

7 0

9/10 is equivalent to 90/100 and the answer under mine or above or the other answer should be reported. They got points for a question they did not answer correctly.

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A normal population has a mean of 12.2 and a standard deviation of 2.5. compute the z value associated with 14.3. what proportio

Citrus2011 [14]

The z-score associated with 14.3 is 0.84. 0.2995 of the population is between 12.2 and 14.3. 0.1894 of the population is less than 10.0.

The formula for a z-score is

z=(X-μ)/σ

With our data, we have:
z=(14.3-12.2)/2.5=0.84

The z-score associated with the mean is 0.5. To find the proportion of the population between the mean and 14.3, subtract 0.7995 (the proportion of population below the z-score of 0.84, using http://www.z-table.com) and 0.5:

0.7995 - 0.5 = 0.2995.

The z-score for 10.0 is

(10.0-12.2)/2.5 = -0.88. The proportion of the population less than this is 0.1894.

5 0

3 years ago

write the ratio for the following description. kaleel made three times as many baskets as john during basketball practice.

Elena L [17]

Ratio is 3:1

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5 0

3 years ago

Write the ratio for cosB

natima [27]

Answer:

11/61

Step-by-step explanation:

cos b = adj side/hypotenuse

=11/61

4 0

2 years ago

Read 2 more answers

he has 1/6 of the pie left over. if he shares it with three friends, how much of original pie will each of the four people get?

mario62 [17]

Sharing with three friends:
1/24 of original pie
Sharing with little brother:
1/12 of original pie

3 0

3 years ago

Read 2 more answers

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.

4 0

3 years ago