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3 years ago

What is the equation of the line that passe throught the point (4, -6) and has a slope of -3

Mathematics

2 answers:

JulijaS [17]3 years ago

7 0

Answer:

your equation should look like:

y+6= -3(x-4)

ipn [44]3 years ago

3 0

Answer:

y=-3x+6

Step-by-step explanation:

y=mx+b

y=-3x+b

shift the parent function y=-3x to where it crosses the point and the y intercept is 6. Put that into your equation for the final answer

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A sock drawer contains eight navy blue socks and five black socks with no other socks. If you reach in the drawer and take two s

Rzqust [24]

Answer:

a. the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b. the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c. the probability of either 2 navy socks is picked or one black & one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

Step-by-step explanation:

A sock drawer contains 8 navy blue socks and 5 black socks with no other socks.

If you reach in the drawer and take two socks without looking and without replacement, what is the probability that:

Solution:

total socks = N = 8 + 5 + 0 = 13

a) you will pick a navy sock and a black sock?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick the black sock, total number of socks left is 12.

Let B be the probability of picking a black sock again.

P (B) = 5/12.

Then, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b) the colors of the two socks will match?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick another navy sock, total number of socks left is 12.

Let B be the probability of another navy sock again.

P (B) = 7/12.

Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

Let D be the probability of picking a black socks first.

Then P (D) = 5/13

without replacing the black sock, will pick another black sock, total number of socks left is 12.

Let E be the probability of another black sock again.

P (E) = 4/12.

Then, the probability of picking 2 black sock = P (D & E)

= (5/13 ) * (4/12) = 5/39 = 0.128

Now, the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c) at least one navy sock will be selected?

this means, is either you pick one navy sock and one black or two navy socks.

so, if you will pick a navy sock and a black sock, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

also, if you will pick 2 navy sock, Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

now either 2 navy socks is picked or one black one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

4 0

3 years ago

6= 5r - 14<br> What are the steps you would use to solve the equation?

gregori [183]

6 = 5r - 14
6 + 14 = 5r
20 = 5r
r = 4

7 0

3 years ago

How many years will it take James to recover his investment?

My name is Ann [436]

I don’t think anyone knows this one not even James

6 0

3 years ago

Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is