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3 years ago

6

Robert has 4 times as many pennies as he does nickels. If Robert has $1.44 worth of pennies and nickels, how many nickels does h

e have?

1 answer:

Nostrana [21]3 years ago

6 0

He is going to have 21 nickels

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Find the modulus and the unit vector along the sum of the vectors. - r₁ +r₂ +r3 given that r₁ = 2i+j+4k - r₂ = 3i-2j+7k, r3 = 5i

bazaltina [42]

j+8k

first -2i+j+4k-3i-2j+7k+5i+2j-3k

8 0

2 years ago

What is the probability of drawing the compliment of a king or a

inna [77]

Answer:

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Let 'S' be the sample space associated with the drawing of a card

n (S) = 52C₁ = 52

Let E₁ be the event of the card drawn being a king

$n( E_{1} ) = 4 _{C_{1} } = 4$

Let E₂ be the event of the card drawn being a queen

$n( E_{2} ) = 4 _{C_{1} } = 4$

But E₁ and E₂ are mutually exclusive events

since E₁ U E₂ is the event of drawing a king or a queen

<u><em>step(ii):-</em></u>

The probability of drawing of a king or a queen from a standard deck of playing cards

P( E₁ U E₂ ) = P(E₁) +P(E₂)

$= \frac{4}{52} + \frac{4}{52}$

P( E₁ U E₂ ) = $\frac{8}{52}$

<u><em>step(iii):-</em></u>

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards

$P(E_{1}UE_{2}) ^{-} = 1- P(E_{1} U E_{2} )$

$P(E_{1}UE_{2}) ^{-} = 1- \frac{8}{52}$

$P(E_{1}UE_{2}) ^{-} = \frac{52-8}{52} = \frac{44}{52} = 0.846$

<u><em>Conclusion</em></u>:-

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846

5 0

3 years ago

Now go to diagonals of parallelograms. You'll see two line segments, and marked with their midpoints, E and F. Verify that E and

AURORKA [14]

Answer:

AE=4

EC=4

BF=2.24

FD=2.24

Step-by-step explanation:

here we go honey

5 0

3 years ago

Solving the system of linear equations by graphing

RideAnS [48]

A) y = 2x -4
B) x + 2y = 12

A) -2x + y = -4 then multiplying Equation B by 2
B) 2x + 4y = 24

Adding the equations
5y = 20
y = 4
x = 4
Okay, that was solved by elimination.
If these equations were graphed, they would be two lines that would meet at the point (4, 4)

6 0

3 years ago

Laurie and Maya sold at most $50 worth of get-well and friendship cards. The friendship cards, x, were sold for $2 each and the

stellarik [79]

Answer:

The point is (15,10)

Step-by-step explanation:

Given data:

cost of card $50

x card sold for $2

y card sold for $1.50

from the above information given above we have following relation

2x + 1.50y = 50

the above equation is linear, therefore to find the point to point out the number of card sold, hit and try method should be used. select the proper point (x, y) so that after putting in above equation there result must not be greater than 50

Try for (15,10)

$2 \times 15 + 1.50\times 10 \geq 50$

$45 \geq 50$

4 0

4 years ago