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riadik2000 [5.3K]
3 years ago
10

Can somebody help me out my grade are bad

Mathematics
2 answers:
andreyandreev [35.5K]3 years ago
8 0

Answer:

3, 2, 6

Step-by-step explanation:

from left, top, and bottom

dimulka [17.4K]3 years ago
5 0

Answer:

2/6

Step-by-step explanation:

Convert 1/2 to something similar to 1/6

1 times 3 is 3

2 times 3 is 6

so now do 3/6 -1/6

Dont change the denominator at all, just subtract the numerator. By doing that you get 2/6

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Please help! Here's the question. I really need it.
Korvikt [17]
Let’s first cut/separate this figure into two shapes: rectangle and a triangle

The area of the rectangle: 9 • 2 = 18
The area of the triangle: 1/2 • 8 • 4 = 16
The total area: 18 + 16 = 34
He needs 34 m square of grass
5 0
3 years ago
No links i will give you brainliest. please
CaHeK987 [17]

Answer:

n=radical of 28

Step-by-step explanation:

n^2=bc

n^2=28

n=radical of 28

8 0
3 years ago
When m and I are cut by transversal n. if m<3=67° , determine m<6.​
marshall27 [118]

Answer:

  • 113°

Step-by-step explanation:

Angles 3 and 6 are consecutive interior angles

<u>As per definition, consecutive internal angles add to 180°</u>

  • m∠3 = 67° ⇒
  • m∠6 = 180° - 67° = 113°
3 0
3 years ago
Trapezoid RSTV∼trapezoid WXYZ .
ICE Princess25 [194]
WX/RS = (28 ft)/(40 ft) = 7/10

Selection D is appropriate.
5 0
3 years ago
Read 2 more answers
A highway engineer knows that his crew can lay 5 miles of highway on a clear day, 2 miles on a rainy day, and only 1 mile on a s
lesya692 [45]

Answer:

Mean = 3.7

Variance = 2.61

Step-by-step explanation:

From the data given; we can represent our table into table format for easier solution and better understanding.

Given that:

A highway engineer knows that his crew can lay 5 miles of highway on a clear day, 2 miles on a rainy day, and only 1 mile on a snowy day

Let X represent the crew;

P(X) represent their respective probabilities

                   clear  day           rainy day            snowy day

X                  5                          2                           1

P(X)              0.6                       0.3                       0.1

From Above; we can determine our X*P(X) and X²P(X)

Let have the two additional columns to table ; we have

X                      P(X)                      X*P(X)                       X²P(X)

5                        0.6                          3                               15

2                        0.3                        0.6                             1.2

1                         0.1                          0.1                            0.1

Total                  1.0                        3.7                             16.3

The mean \mu can be calculated by using the formula:

\sum \limits ^n _{i=1}X_i P(X_i)

Therefore ; mean \mu = 3.7

Variance \sigma^2 = \sum \limits ^n _{i=1}X^2_i P(X_i)- \mu^2

Variance = 16.3 -3.7²

Variance = 16.3 - 13.69

Variance = 2.61

4 0
4 years ago
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