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3 years ago

1

Mathematics

2 answers:

Assoli18 [71]3 years ago

8 0

Answer:

Area of a triangle= b · h/2

b) 10 is the correct answer because 10 x 3/2 = 15

const2013 [10]3 years ago

7 0

A) 12 cm ..............

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Find the values of m and c if the straight line y=mx+c passes through the point (-2,5) and has a gradient of 4.

MAVERICK [17]

(2,5) and (4,13)

y - y1 = y2 - y1

x - x1 x2 - x1

y - 5 = 13 - 5 = 8/2 = 4 x - 2 4 - 2

y - 5 = 4x - 8

y = 4x - 3

m = 4, c = -3

6 0

3 years ago

Which one is right ??

schepotkina [342]

C: Hexagon. It Seems To Be The Logical Choice

6 0

4 years ago

Read 2 more answers

The computers of six faculty members in a certain department are to be re- placed. two of the faculty members have selected lapt

FinnZ [79.3K]

Answer:

A) The probability that both selected setups are for laptop computers is 0.067

B)The probability that both selected setups are desktop machines is 0.4

C)The probability that at least one selected setup is for a desktop computer is 0.933

D)The probability that at least on computer of each type is chosen for setup is 0.533

Step-by-step explanation:

Number of laptops = 2

Number of desktops = 4

Total number of outcomes = 15

a) what is the probability that both selected setups are for laptop computers?

Total number of outcomes = 15

So, the probability that both selected setups are for laptop computers = $\frac{1}{15}=0.067$

b)what is the probability that both selected setups are desktop machines?

Number of desktops = 4

Number of desktops to be chosen = 4

We will use combination

No. of ways to select two desktops = $^4C_2=\frac{4!}{2!(4-2)!}=6$

So,the probability that both selected setups are desktop machines= $\frac{6}{15}=0.4$

P(at least 1 desktop)=1-P(No desktop)

P(at least 1 desktop)=1-P(both laptops)

P(at least 1 desktop)=1-0.067=0.933

So,the probability that at least one selected setup is for a desktop computer is 0.933

d) what is the probability that at least on computer of each type is chosen for setup?

No. of ways to select one desktop $=^4C_1=\frac{4!}{1!(4-1)!}=4$

No. of ways to select one laptop = $^2C_1=\frac{2!}{1!(2-1)!}=2$

So, No. of ways to select one laptop and one desktop= $4 \times 2 = 8$

So,the probability that at least on computer of each type is chosen for setup= $\frac{8}{15}=0.533$

6 0

3 years ago

What is the full replacement cost of a new floor plan if the buliding cost $35 per square

stira [4]

What's the measurements of the floorplan

7 0

4 years ago

62.837 expanded for decimal

Liono4ka [1.6K]

Answer:

60 + 2 + 0.8 +0.03 +0.007

Step-by-step explanation:

Add these up and yo get the same number: 62.837.

8 0

3 years ago