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ddd [48]
3 years ago
9

10 2/5 x 15 5/6 and 1 1/3 x 1 1/3

Mathematics
1 answer:
Komok [63]3 years ago
5 0
10 2/5 x 15 5/6 = 164.632            11/3 x 11/3 = 13.2
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If two of these shapes are randomly chosen, one after the other without replacement, what is the probability that the first will
aleksandrvk [35]

Answer:

P(T\ n\ S) = \frac{1}{14}

Step-by-step explanation:

*Missing Part of the Question*

4 stars

5 triangles

3 circles

3 squares

Required

Determine the probability of triangle being first then square being second

Total = 4 + 5 + 3 + 3

Total = 15

Represent the triangle with T and square with S

So, we're solving for P(T n S)

P(T\ n\ S) = P(T) * P(S)

Solving for P(T)

P(T) = \frac{n(T)}{Total}

P(T) = \frac{5}{15}

Solving for P(S)

The question implies a probability without replacement;

Hence Total has now been reduced by 1

Total = 14

P(S) = \frac{n(S)}{Total}

P(S) = \frac{3}{14}

Recall that

P(T\ n\ S) = P(T) * P(S)

P(T\ n\ S) = \frac{5}{15} * \frac{3}{14}

P(T\ n\ S) = \frac{15}{15 * 14}

P(T\ n\ S) = \frac{1}{14}

Hence, the required probability is

P(T\ n\ S) = \frac{1}{14}

3 0
3 years ago
Use properties of addition and subtraction to evaluate the expression <br> -42 - 29 - 28
Setler [38]
(-42) (-29) (-28)
-42 - 29 = -71
-71 - 28 = -99

-99 is your answer

hope this helps
7 0
3 years ago
Help plzzz will give brainliest
Rasek [7]

Answer:

the amswer is 5 hope this helps you

3 0
3 years ago
PLEASE HELP ME ASAP???!!!!!!!!
Ivenika [448]

Answer:

I'm very very sorry I cannot help you I wish I could help sorry bud.

8 0
3 years ago
If I cut a piece that was 5 centimeters wide and had an area of 20 centimeters. How long was the piece?
Soloha48 [4]
Area= Length x width
So....
Length= Area divided by width.
Plug in equation
20 divided by 5= L
L= 4
Answer is 4 cm
6 0
3 years ago
Read 2 more answers
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