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3 years ago

The ratio of coloured paper the white paper is 3:8 if there are 48 white paper how many coloured paper are there

Mathematics

1 answer:

Nuetrik [128]3 years ago

7 0

Answer:

Step-by-step explanation:

Calculation to determine how many coloured paper are there

Let 3x represent the number of coloured papers

Let 8x represent the number of white papers

Given that 8x=48

Hence,

x=48/8

x=6

Now let determine how many coloured paper are there

Coloured paper=3x

Coloured paper=3(6)

Coloured paper=18

Thereffore the number of coloured paper that are there is 18

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Which function has the greatest maximum y-value?

il63 [147K]

Answer:

F(x) and h(x)

Step-by-step explanation:

f(x) has a high of three and so does H(x), g(x) has a high of -2 so it doesn't count

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3 years ago

I Have 10 Minutes Hurry for Brainlist and pls be accurate

yulyashka [42]

To find the equation we will put the values of x and y in the equations and then check if LHS=RHS.

$\large\red{\sf{y=3x-1}}$

$\begin{gathered}\\ \sf\longmapsto 9=3(1)-1\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9=3-1\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9≠2 \end{gathered}$

$\large\green{\sf{y=5x+4}}$

$\begin{gathered}\\ \sf\longmapsto 9=5(1)+4\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9=5+4\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9=9\end{gathered}$

$\large\blue{\sf{y=-2x-8}}$

$\begin{gathered}\\ \sf\longmapsto 9=-2(1)-8\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9=-2-8\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9≠10\end{gathered}$

$\large\pink{\sf{y=6x-8}}$

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Hence, Option B. y=5x+4 satisfies the system (1,9)

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2 years ago

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sveticcg [70]

Answer:

Step-by-step explanation:

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3 years ago

How much money invested at 5% compounded continuously for 3 years will result in $820

nikitadnepr [17]

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3 years ago

A bag contain 3 black balls and 2 white balls.

Troyanec [42]

Answer:

Step-by-step explanation:

Total number of balls = 3 + 2 = 5

$Probability \ of \ taking \ 2 \ black \ ball \ with \ replacement\\\\ = \frac{3C_1}{5C_1} \times \frac{3C_1}{5C_1} =\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\\\\$

$Probability \ of \ one \ black \ and \ one\ white \ with \ replacement \\\\= \frac{3C_1}{5C_1} \times \frac{2C_1}{5C_1} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$

Probability of at least one black( means BB or BW or WB)

$=\frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5} \\\\= \frac{9}{25} + \frac{6}{25} + \frac{6}{25}\\\\= \frac{21}{25}$

Probability of at most one black ( means WW or WB or BW)

$=\frac{2}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{2}{5} \times \frac{2}{5} + \frac{3}{5}\\\\= \frac{4}{25} + \frac{6}{25} + \frac{6}{25}\\\\=\frac{16}{25}$

a) Probability both black without replacement

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

b) Probability of one black and one white

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

c) Probability of at least one black ( BB or BW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{6}{20} + \frac{6}{20} \\\\=\frac{18}{20} \\\\=\frac{9}{10}$

d) Probability of at most one black ( BW or WW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{2}{20} + \frac{6}{20} \\\\=\frac{14}{20}\\\\=\frac{7}{10}$

6 0

3 years ago