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3 years ago

Refer the homework helper what is the diffrence between the greatest green bean weight and the least green bean weight?

Mathematics

1 answer:

QveST [7]3 years ago

6 0

Answer:

3/4 pounds

Step-by-step explanation:

Given the weights :

weight of beans 1/4 pound, 1/2 pound, 3/4. pound and 1 pound.

Greatest weight = 1 pound

Least weight = 1/4 pound

Difference = greatest weight - Least weight

Difference = (1 - 1/4) pound

Difference in weight = 3/4 pounds

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3 years ago

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3 years ago

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malfutka [58]

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4 years ago

If<br>A = i +3j -2K and B=4i-2j+4k<br>Find (2A+B).(A-2B)

Marizza181 [45]

GIVEN :

a = 1/√10 ( 3i + k) and

b = 1/7 ( 2i + 3j - 6k)

TO FIND :

( 2a- b) . [ ( a x b ) x ( a + 2b)]

SOLUTION :

◆Going with the equation given,

( 2a - b) . [ ( a x b ) x ( a + 2b)]

= (2a - b) [( a×b×a) + 2(a×b)×b]

◆BAC - CAB RULE,

A×B×C = B( A.B ) - C(A.B )

= (2a- b ) [ (b (a.a ) - a (a.b ) + 2b ( a.b) -2b (a.b]

Solving further

= (2a - b )(b - 2a)

= -4a.a -b.b

=-5.

Answer:

( 2 - b) . [ ( a x b ) x ( a + 2b)] = -5

Hoped I helped

8 0

3 years ago

A container contains 12 diesel engines. The company chooses 5 engines at random, and will not ship the container if any of the

Rufina [12.5K]

Answer: $\frac{^{10}C_5}{^{12}C_5}$

Step-by-step explanation:

Given a total of 12 diesel engines

out of which 2 are defective

Company have to choose 5 good engines to ship.

Therefore no of ways in good engines are selected is $^{10}C_5$

no of ways in which any 5 engines are selected from a total of 12 engines is

$^{12}C_5$

Therefore the required probability = $\frac{^{10}C_5}{^{12}C_5}$

= $\frac{7}{22}$ =0.318

3 0

3 years ago