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2 years ago

In a bag, there are 25 marbles. 8 are red and the rest are green. If there are 100 marbles in. a bag, how many are red? *

Mathematics

1 answer:

Sophie [7]2 years ago

4 0

Answer:

32 marbles are red.

Step-by-step explanation:

Think of a quarter, 4 of them equal a dollar.

25 x 4 = 100

So, you will multiply 8 by 4.

8 x 4 = 32

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What is the missing part

GrogVix [38]

Do 24 times 25 and then divide that in half

7 0

2 years ago

Can someone help me with this? I don't understand how to solve it.<br> (3x^2a - 4y^az^3a)^2

EastWind [94]

(3x²ᵃ - 4yᵃz³ᵃ)² =

= 9x⁴ᵃ - 12x²ᵃyᵃz³ᵃ + 16y²ᵃz6ᵃ

6 0

2 years ago

The Blue Diamond Company advertises that their nut mix contains (by weight) 40% cashews, 15% Brazil nuts, 20% almonds and only 2

jek_recluse [69]

Answer:

At the 0.05 level of significance, the claim made by Blue Diamond is true

Step-by-step explanation:

Given that the Blue Diamond Company advertises that their nut mix contains (by weight) 40% cashews, 15% Brazil nuts, 20% almonds and only 25% peanuts.

$H_0: The mix are as per percents given\\H_a:Atleast two differ from the above percent.$

(Two tailed chi square test at 5% significance level)

Chi square is calculated as (Obs_exp)^2/Expected

Cashews Brazil nuts Almonds Peanuts Total

Observed 5 6 6 3 20

Expected 40% 15% 20% 25% 1

Expected 8 3 4 5 20

Chi square 1.125 3 1 0.8 5.925

df = 3

p value = 0.1153

Since p value >0.0 we accept null hypothesis.

4 0

3 years ago

How to know how to solve properties of parallelograms

Archy [21]

Answer:

Opposite sides are corresponding (AB = DC).

Opposite angles are corresponding (D = B).

Consecutive angles are supplementary (A + D = 180°).

If one angle is right, then all angles are right.

The diagonals of a parallelogram hemisect each other.

Step-by-step explanation:

5 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago