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Elden [556K]
2 years ago
7

A school has 1000 students. of these, 3/4 are boys. how many of the students are boys? how many of the students are girls?

Mathematics
2 answers:
Whitepunk [10]2 years ago
8 0
750 are boys 1/4 are girls. 3/4 of 1000 is 750 and the remaining 1/4 are girls
weqwewe [10]2 years ago
6 0

Answer:

750 students are boys, 250 are girls.

Step-by-step explanation:

1/4 of 1000 is 250.

250 + 250 + 250 = 750 boys.

if 3/4 are boys, then 1/4 (the rest of the school population) are girls.

150 students are girls.

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marta [7]
I’m not that smart... but let me use first as example.. 240 divided by 4
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3 years ago
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Someone plz help me
topjm [15]

x = 375 and y = 6300. Option C.

Step-by-step explanation:

From the given data we need to find the value of x and y.

First, we will find out the ratio of area and the force.

\frac{Area}{Force}  = \frac{125}{1875} =\frac{150}{2250} =\frac{175}{2625} = \frac{1}{15}

So,

\frac{x}{5625} = \frac{1}{15}

or, x = \frac{5625}{15} = 375

Again,

\frac{420}{y} = \frac{1}{15}

or, y = 420×15 = 6300

Hence,

x = 375 and y = 6300.

7 0
3 years ago
In ΔKLM, l = 48 inches, ∠K=55° and ∠L=46°. Find the length of k, to the nearest inch.\
s2008m [1.1K]

Given:

In  ΔKLM, l = 48 inches, ∠K=55° and ∠L=46°.

To find:

The length of k, to the nearest inch.

Solution:

According to Law of sine,

\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}

In ΔKLM, using law of sine, we get

\dfrac{k}{\sin K}=\dfrac{l}{\sin L}

\dfrac{k}{\sin (55^\circ)}=\dfrac{48}{\sin (46^\circ)}

k=\dfrac{48\times \sin (55^\circ)}{\sin (46^\circ)}

On further simplification, we get

k=\dfrac{48\times 0.819152}{0.7193398}

k=\dfrac{39.319296}{0.7193398}

k=54.66025


Approximate the value to the nearest inch.

k\approx 55


Therefore, the length of k is 55 inch.

4 0
3 years ago
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marusya05 [52]

The Lagrangian is

L(x,y,\lambda)=x+8y+\lambda(x^2+y^2-4)

It has critical points where the first order derivatives vanish:

L_x=1+2\lambda x=0\implies\lambda=-\dfrac1{2x}

L_y=8+2\lambda y=0\implies\lambda=-\dfrac4y

L_\lambda=x^2+y^2-4=0

From the first two equations we get

-\dfrac1{2x}=-\dfrac4y\implies y=8x

Then

x^2+y^2=65x^2=4\implies x=\pm\dfrac2{\sqrt{65}}\implies y=\pm\dfrac{16}{\sqrt{65}}

At these critical points, we have

f\left(\dfrac2{\sqrt{65}},\dfrac{16}{\sqrt{65}}\right)=2\sqrt{65}\approx16.125 (maximum)

f\left(-\dfrac2{\sqrt{65}},-\dfrac{16}{\sqrt{65}}\right)=-2\sqrt{65}\approx-16.125 (minimum)

5 0
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Klio2033 [76]

Answer:4.6h-2.9d-16

Step-by-step explanation:

7 0
3 years ago
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