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FromTheMoon [43]

4 years ago

6

Am man has 6 black socks and 8 white socks in a drawer unpaired what is the probability in the dark the man took out two socks o

f the same color?

1 answer:

lord [1]4 years ago

8 0

See attached diagram of working.

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The value of a car purchased for $20,000 decreases at a rate of 12% per year. What will be the value of the car after 3 years?

timurjin [86]

Answer:

$13,629.44

Step-by-step explanation:

Value of car after 3 years is:

V = $20,000(1.00-0.12)^3, or

$20,000(0.88)^3 = $13,629.44

5 0

3 years ago

Please help -2(x+10)

MissTica

Answer:

-2x-5

Step-by-step explanation:

By distributing the -2 to the x+10 we get, "-2x-5". This is our final simplified answer.

3 0

3 years ago

Read 2 more answers

Which number is equivalent to 1/9?

atroni [7]

5/45, 6/54, 7/63
Any 2 numbers that you divide and have the ratio 1:9

3 0

3 years ago

At a liberal arts college, 90% of the freshmen are enrolled in English 105, 80% are enrolled in Mathematics 101 and 5% are enrol

vova2212 [387]

Answer:

a) There is a 75% probability that the freshman is enrolled in both English 105 and Mathematics 101.

b) There is a 15% probability that the freshman is enrolled in English 105 but not in Mathematics 101.

c) 93.75% probability that the freshman is also enrolled in English 105

Step-by-step explanation:

We solve this problem building the Venn's diagram of these percentages.

I am going to say that:

A are those freshmen enrolled in English 105.

B are those freshman Enrolled in Mathematics 101.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a freshmen is enrolled in English 105 but not in Mathematics 101 and $A \cap B$ is the probability that a freshmen is enrolled in both these classes.

By the same logic, we have that:

$B = b + (A \cap B)$

Solution

90% of the freshmen are enrolled in English 105

So A = 0.9

80% are enrolled in Mathematics 101

So B = 0.8

5% are enrolled in Mathematics 101 but not in English 105.

So b = 0.05

a. What is the probability that the freshman is enrolled in both English 105 and Mathematics 101?

This is $A \cap B$ .

$B = b + (A \cap B)$

$0.8 = 0.05 + (A \cap B)$

$A \cap B = 0.75$

There is a 75% probability that the freshman is enrolled in both English 105 and Mathematics 101.

b. What is the probability that the freshman is enrolled in English 105 but not in Mathematics 101?

This is a.

$A = a + (A \cap B)$

$0.90 = a + 0.75$

$a = 0.15$

There is a 15% probability that the freshman is enrolled in English 105 but not in Mathematics 101.

c. Suppose the freshman chosen is known to be enrolled in Mathematics 101. What is the probability that the freshman is also enrolled in English 105?

This is $P(A \cap B)$ divided by P(B). So 0.75/0.8 = 0.9375.

There is a 93.75% probability that the freshman is also enrolled in English 105

6 0

4 years ago

(4,5) (0,8)<br> Slope intercept form

laila [671]

$(\stackrel{x_1}{4}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{8}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{8}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{4}}}\implies \cfrac{3}{-4}\implies -\cfrac{3}{4}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{-\cfrac{3}{4}}(x-\stackrel{x_1}{4}) \\\\\\ y-5=-\cfrac{3}{4}x+3\implies y=-\cfrac{3}{4}x+8$

7 0

3 years ago