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Verizon [17]
3 years ago
5

A jar of marbles contains the following: two purple marbles, four white marbles, three blue marbles, and two green marbles. What

is the probability of selecting a white marble from a jar of marbles?
Mathematics
1 answer:
Llana [10]3 years ago
4 0

Answer:

4/11

Step-by-step explanation:

Total number of marbles = 2(purple) + 4(white) + 3(blue) + 2(green)

                                          = 11

Number of white marbles = 4

Probability of selecting a white marble =

         number of white marbles/total number of marbles in the jar

         = 4/11

You might be interested in
find the distance between city A and city B if there 15 cm apart on the map with a scale of 2 cm: 16 km
IRISSAK [1]

Answer:

120 km

Step-by-step explanation:

15/2=7.5

7.5 km x 16= 120km

Have a great day!

7 0
3 years ago
kofi and kojo were given 380 to share .kojo had 75 more than kofi. If kofi share is x find an expression for kojo share​
AleksAgata [21]

Answer:

227.5

Step-by-step explanation:

x=Kofi

x+75=Kojo

   x+x+75=380

    2x+75=380

          2x=305

             x=152.5

       Kojo=152.5 plus 75

               227.5

4 0
3 years ago
Hiiii.. please help me with this limit question ​
Alenkasestr [34]

Answer:

π

Step-by-step explanation:

Solving without L'Hopital's rule:

lim(x→0) sin(π cos²x) / x²

Use Pythagorean identity:

lim(x→0) sin(π (1 − sin²x)) / x²

lim(x→0) sin(π − π sin²x) / x²

Use angle difference formula:

lim(x→0) [ sin(π) cos(-π sin²x) − cos(π) sin(-π sin²x) ] / x²

lim(x→0) -sin(-π sin²x) / x²

Use angle reflection formula:

lim(x→0) sin(π sin²x) / x²

Now we multiply by π sin²x / π sin²x.

lim(x→0) [ sin(π sin²x) / x² ] (π sin²x / π sin²x)

lim(x→0) [ sin(π sin²x) / π sin²x] (π sin²x / x²)

lim(x→0) [ sin(π sin²x) / π sin²x] lim(x→0) (π sin²x / x²)

π lim(x→0) [ sin(π sin²x) / π sin²x] [lim(x→0) (sin x / x)]²

Use identity lim(u→0) (sin u / u) = 1.

π (1) (1)²

π

Solving with L'Hopital's rule:

If we plug in x = 0, the limit evaluates to 0/0.  So using L'Hopital's rule:

lim(x→0) [ cos(π cos²x) (-2π cos x sin x) ] / 2x

lim(x→0) [ -π cos(π cos²x) sin(2x) ] / 2x

-π/2 lim(x→0) [ cos(π cos²x) sin(2x) ] / x

Again, the limit evaluates to 0/0.  So using L'Hopital's rule one more time:

-π/2 lim(x→0) [ cos(π cos²x) (2 cos(2x)) + (-sin(π cos²x) (-2π cos x sin x)) sin(2x) ] / 1

-π/2 lim(x→0) [ 2 cos(π cos²x) cos(2x) + π sin(π cos²x) sin²(2x) ]

-π/2 (-2)

π

8 0
3 years ago
What is the value of y? 3(y 5)=–2y3(y 5)=–2y <br> a. –5 <br> b. –5 <br> c. –3 <br> d. –335?
prohojiy [21]
What are the values in (y 5) is it (y+5) or (y-5) i cannot answer until given the full problem.
4 0
3 years ago
3(x + 5) = 39 I dont understand this problem
Dvinal [7]
3(x+5)=39
3x+15=39
3x=24
x=8
5 0
3 years ago
Read 2 more answers
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