El primer día de la tormenta de nieve hubo 9,2 centímetros de nieve. Durante el segundo día de la tormenta, cayeron otros 18,2 centímetros. Si la nevada total durante la tormenta de nieve de tres días fue de 39,1 centímetros, ¿cuánta nieve cayó el tercer día?​

Answers

Answer 1

Answer:

11.7

Step-by-step explanation:

39.1 - 9.2 = 29.9

29.9 - 18.2 = 11.7


Related Questions

f(x) = 3x + 10
х
f(x)
-3
-2
-1
-4

Answers

The answer is 89 you missed a step

Which equation shows the point-slope form of the line that passes through (3, 2) and has a slope of 1/3
O y + 2 =1/3(x + 3)
O y-2=1/3(x-3)
O y + 3 = 1/3(x+ 2)
O y-3= 1/3(x-2)

Answers

Answer:

y - 2 = 1/3(x - 3)

Step-by-step explanation:

Point slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

Plug in the slope:

y - y1 = m(x - x1)

y - y1 = 1/3(x - x1)

Plug in the given point:

y - y1 = 1/3(x - x1)

y - 2 = 1/3(x - 3)

So, the correct answer is y - 2 = 1/3(x - 3)

simplify the expression (2r^4y4xy^2) completely

Answers

Step-by-step explanation:

we can simplify the stuff inside the parentheses to

[tex] \frac{ {x}^{3} }{2y} [/tex]

now we need to multiply it with itself, giving us

[tex] \frac{ {x}^{6} }{4 {y}^{2} } [/tex]

so yeah, D is the correct answer

Bill can hit a bucket of 323 golf balls in 17 hours.
How many golf balls can Bill hit in 23 hours?

Answers

Bill can hit 437 golf balls in 23 hours.

Explanation:

Bill’s unit rate of golf balls per hour is:
323 ÷ 17 = 19

19 golf balls per hour

To find how many golf balls Bill can hit in 23 hours, multiply:
19 • 23 = 437

Therefore, Bill can hit 437 golf balls in 23 hours.

Please, say if my answer is right
If its, glad to help!

-SpaceMarsh

3. Two dice are rolled. What’s the conditional probability that both dice are 5’s if it’s known that the sum of points is divisible by 5?

Answers

Answer:

[tex]Pr =\frac{1}{3}[/tex]

Step-by-step explanation:

Given

[tex]S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)[/tex]

[tex](3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)[/tex]

[tex](5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)\}[/tex] --- sample space

First, list out all outcomes whose sum is divisible by 5

[tex]A = \{(4,6), (5,5),(6,4)\}[/tex]

So, we have:

[tex]n(A) = 3[/tex]

Next, list out all outcomes that has an outcome of 5 in both rolls

[tex]B = \{(5,5)\}[/tex]

[tex]n(B) =1[/tex]

The required conditional probability is:

[tex]Pr =\frac{n(B)}{n(A)}[/tex]

[tex]Pr =\frac{1}{3}[/tex]

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