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4 years ago

I WILL MARK BRAINLIEST!!!!!!!!!!

Mathematics

2 answers:

soldier1979 [14.2K]4 years ago

7 0

The answer is letter C

solmaris [256]4 years ago

4 0

It would be 12 !
considering you have to go up by two each unit; having to go up 6 boxes, then times two for what its counting by, would give you 12 !

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An investor purchased a house 8 years ago for $48,000 This year it was appraised at $73,500 Complete parts (a) through (d) b

raketka [301]

Answer:

b = $48000, m = $3187.5 / year

Step-by-step explanation:

The equation of a linear function is given as y = mx + b, where m is the rate of change, b is the value of y when x = 0, y = dependent variable and x = dependent variable.

Given that V = mt + b:

b = initial price of the house at 0 years = $48000

V = mt + 48000, At 8 years the house is appraised at $73,500

73500 = 8m + 48000

8m = 73500 - 48000

8m = 25500

m = 3187.5

6 0

4 years ago

Which is a pair of dependent events?

pashok25 [27]

Answer:

Two events are dependent if the outcome of the first event affects the outcome of the second event, so that the probability is changed.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

I'll give you brainliest once i know its right !!

Masteriza [31]

Answer:

Step-by-step explanation:

H77777777

5 0

3 years ago

Read 2 more answers

Examine the following table of points, which are all on a certain line.

Lapatulllka [165]

$\bf \begin{array}{|cc|ll} \cline{1-2} x&y\\ \cline{1-2} {-2}~~\ast&{4}~~\ast\\ 0&2\\ 1&1\\ {3}~~\ast&{-1}~~\ast\\ \cline{1-2} \end{array}~\hspace{10em} (\stackrel{x_1}{-2}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-4}{3-(-2)}\implies \cfrac{-5}{3+2}\implies \cfrac{-5}{5}\implies -1$

4 0

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