Write the equation of a line that would be parallel to the line given. 2 10) y=-2/3x+10

What inequality is graphed below?

vovikov84 [41]

Answer:

X>-6

Step-by-step explanation:

That's the answer.....

7 0

3 years ago

Read 2 more answers

Ken has $950 in a savings account at the beginning of the summer. He wants to have at least $600 in the account by the end of th

Neko [114]

Every week he withdraws $35 from the $950 in his savings, and he wants at least $600 by the end of the summer. So we could write:

$\sf 950-35x\ge 600$

Where 'x' is the number of weeks. Now let's solve it for 'x'.

Subtract 950 to both sides:

$\sf -35x\ge -350$

Divide -35 to both sides, don't forget to switch the sign when dividing/multiplying by a negative number:

$\sf x\le 10$

So he can only withdraw at money for at most 10 weeks if he wants at least $600 left in his account.

6 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

1)

a)

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

b)

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

c)

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

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

d)

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

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

2)

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

Need help with number 4

zmey [24]

–6(x – 2y) when x = –3 and y = –5
so
–6(x – 2y)
= –6(-3 – 2(-5))
= -6(-3 + 10)
= -6(7)
= -42

answer
A. -42

3 0

3 years ago

George and carmen went on a bicycle trip. they took a bus to their starting piont, and then biked the rest. they traveled 275 ki

rodikova [14]

We need to do a system of equations here.

x + y = 275 (if you travel x km by bike and y km by bus, then you travel 275 km as given in the problem)
y = x + 55 (the problem stated that they were bussed (y) the amount they biked plus 55 more km (x + 55))

The second equation is already solved for y. So, we can plug it in to the first equation.

x + (x + 55) = 275
2x + 55 = 275 [combine x terms]
2x = 220 [subtract 55 from both sides]
x = 110 [divide by 2 to isolate x and solve for it]

Now we know that x is 110, the distance they traveled by bike.
And that's what we needed to answer the problem!

5 0

4 years ago