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3 years ago

Help ASAP please, 10 points if you do

Mathematics

2 answers:

yawa3891 [41]3 years ago

8 0

I’d say b.

agree with the dude above

Andre45 [30]3 years ago

7 0

I would say B.

Your starting term is 10, so that is your a1 value. Each time you add 4 to the number before, like 10+4=14 and so on, so that would mean (a-1)+4.

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<img src="https://tex.z-dn.net/?f=5%20%5Ctimes%20%5Cfrac%7B4%7D%7B7%7D%20" id="TexFormula1" title="5 \times \frac{4}{7} " alt="5

Butoxors [25]

$\bf 5\times \cfrac{4}{7}\implies \cfrac{5\cdot 4}{7}\implies \cfrac{20}{7}\implies \cfrac{14+6}{7}\implies \cfrac{14}{7}+\cfrac{6}{7}\implies 2+\cfrac{6}{7}\implies 2\frac{6}{7}$

4 0

3 years ago

Read 2 more answers

PLEASE HELP The points (6, 8) and (x, 12) are two points on a graph of an inverse variation. Find the missing value. Show all wo

UNO [17]

Answer:

X=9

Step-by-step explanation:

6/8=3/4

x/12=3/4

x=9

6 0

4 years ago

Name and find the measure of each angle in the figure. <br><br><br> HELP PLEASE

stich3 [128]

Is the algebra, if so what type?

4 0

3 years ago

A piece of blue yarn is 5/ 3/7 yards long. A piece of pink yarn is 5 times as long as the blue yarn. How much longer is the pink

MaRussiya [10]

Answer:

27 1/7

Step-by-step explanation:

Multiply 5×5=25 and afterwards multiply 3/7×5=2 1/7

8 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