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Lelu [443]
2 years ago
10

Jordan went to shelter to adopt a dog. There were 10 brown dogs and 10 black dogs. Jordan picked a dog, went home, then decided

he wanted to go back and adopt another dog. What is the probability that he picked two black dogs?
Mathematics
2 answers:
mixas84 [53]2 years ago
5 0

10/38 is what i think it is

amid [387]2 years ago
3 0

Answer:

The probability is less likely and 2/20

Please mark me as brainliest.

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How many pounds of candy that cells for $0.66 per Ib must be mixed with candy that sells for $1.31 per Ib to obtain 9 Ib of a mi
faltersainse [42]

Answer:  \frac{306}{65} pounds

Step-by-step explanation:

Let x pounds of $ 0.66 per lb candies mixed with y pounds of $ 1.31 per lb candies to obtain 9 lb of $ 0.97 per lb candies,

Hence, we can write,

x + y = 9 ------- (1)

And, 0.66 of x + 1.31 of y = 0.97 of (x + y)

⇒ 66 x + 131 y = 97 (x+y)

⇒ 66 x +131 y - 97 x - 97 y = 0

⇒ -31 x + 34 y = 0  ------(2)

By solving equation (1) and (2)

We get,

x=\frac{306}{65} and y=\frac{279}{65}

Hence, the quantity of 0.66 per lb candies = \frac{306}{65}  pounds

7 0
3 years ago
Bill bought 38 1/2 feet of wire fencing to put around his blueberry plants. He cut the wire into several pieces, each one measur
Nadya [2.5K]

Answer: 7

Step-by-step explanation:

You are given that Bill bought 38 1/2 feet of wire. Convert the mixed fraction into improper fraction.

The length L1 = 77/2

You are given that he cut the wire into several pieces, each one measuring 5 1/2 feet. Convert it too to improper fraction.

The length L2 = 11/2

To know how many number of pieces of wire he have after making the cuts, divide length L1 by length L2. That is,

L1/L2 = 77/2 ÷ 11/2

Change the division to multiplication by reciprocating 11/2 to 2/11

77/2 × 2/11 = 77/11 = 7

Therefore, he had 7 pieces of wires after making the cuts.

7 0
3 years ago
Prepare the journal entry to record the issuance of 5,000 shares of $20 par common stock for $25 per share
vagabundo [1.1K]

Answer:

value preferred stock for $65,250 cash

<h2> HOPE THIS HELP PLEASE TOUCH THANKS.</h2>

7 0
2 years ago
Suppose the equation ax^2+bx+c=0 has no real solution and a graph of the related function has a vertex that lies in the second q
Veseljchak [2.6K]
If it has no real solutions, that means the graph does not intersect the x axis

since we have ax^2+bx+c=0, the parabola opens either up or down
since the vertex is in the second quadrant (x is negative and y is positive in this reigon) and the graph does not cross the x axis, the parabola must open up
if the value of 'a' is positive, then the parabola opens up

so 'a' must be positive



if it is translated to the 4th quadrant, then the vertex is now below the x axis
it will now have 2 x intercepts because the vertex is in the 4th quadrant and look at a graph of a parabola opening up with vertex in 4th quadrant and seehow many time it crosses the x axis
7 0
3 years ago
PleasePICK ON OF THE MUTIPL CHOICE<br> PLESE AND<br> THANNNNKKKSSS!!!!
erica [24]

it is infintely many  i  think so


4 0
3 years ago
Read 2 more answers
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