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Karolina [17]
3 years ago
5

A fish store received a shipment of 246 goldfish. The owner of the store arranged the fish into small tanks of 8 fish per tank.

How many fish were left over?
Mathematics
2 answers:
Kay [80]3 years ago
7 0

Answer:

6

Step-by-step explanation:

saw5 [17]3 years ago
6 0

Answer:

he would need 31 tanks

30 tanks would hold 240 fish with 6 left over

Step-by-step explanation:

246/8= 30.75

30*8 = 240

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Which point is on the graph of f(x) = 2 • 5x?
Inessa [10]

Answer:

1, 10

Step-by-step explanation:

6 0
2 years ago
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A triangle has ∠A, ∠B, and ∠C.<br> ∠B is 20° more than ∠A.<br> ∠C is double ∠B.<br> How big is ∠A?
san4es73 [151]

Answer:

30°

Step-by-step explanation:

∠A + ∠B + ∠C = 180

∠B = ∠A + 20

∠C = 2 * ∠B

<A + (∠A + 20) + (2 * ∠B) = 180

<A + (∠A + 20) + (2 * (∠A + 20)) = 180

<A + (∠A + 20) + 2∠A + 40 = 180

4∠A + 60 = 180

4∠A = 120

∠A = 30

5 0
3 years ago
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Oliver climbed 222 meters of a steep mountain reach the summit. He then rappelled down the other side of the mountain to return
nikdorinn [45]
B -222

Because he fell/descended 222 meters. And falling or going down produces a negative integer. Also, if you were actually doing the math to try to find where Oliver is currently located, you wouldn't add 222, or 0, or subtract -444. You would subtract 222. 

In addition, be aware that the question is asking you an integer to represent a situation, not the answer of where Oliver is. 

6 0
2 years ago
The manager of a music store has kept records of
Marysya12 [62]

Answer:

(a) P(x\le 3) = 0.75

(b) P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

(c) P(x \ge 5) = 0.10

(d) P(x = 1\ or\ 2) = 0.55

(e) P(x > 2) = 0.45

Step-by-step explanation:

Given

\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

Solving (b): Probability of at most 3 CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

<em />

Solving (c): Probability of 5 or more CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x \ge 5) = P(5) + P(6\ or\ more)

This gives:

P(x\ge 5) = 0.05 + 0.05

P(x \ge 5) = 0.10

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}

This probability is calculated as:

P(x = 1\ or\ 2) = P(1) + P(2)

This gives:

P(x = 1\ or\ 2) = 0.30 + 0.25

P(x = 1\ or\ 2) = 0.55

Solving (e): Probability of more than 2 CDs

Here, we consider:

\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)

This gives:

P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05

P(x > 2) = 0.45

3 0
3 years ago
Right triangle ABC is similar to right triangle DEF. If the side lengths for triangle ABC are 15, 20, 25, respectively, which va
solniwko [45]

Answer:

A = 15   B = 20   C = 25

D = 30   E = 40   F = 50

D = 3    E = 4     F = 5

Step-by-step explanation:

Given

Similar Triangles: ABC and DEF

A = 15

B = 20

C = 25

Required

Determine the sides of DEF

No options were given, so I will solve on a general terms.

Since both triangles are similar, then the following relationship exists.

DEF = ABC * n

i.e.

D = A * n    E = B * n     F = C * n

Where

n = Scale\ Factor

Assume n = 2.

So, we have:

D = A * n    

D =15 *2

D = 30

E = B * n    

E = 20 * 2

E = 40

F = C * n

F = 25 * 2

F = 50

Assume n = \frac{1}{5}.

So, we have:

D = A * n    

D =15 *\frac{1}{5}

D = 3

E = B * n    

E = 20 * \frac{1}{5}

E = 4

F = C * n

F = 25 * \frac{1}{5}

F = 5

So, the possible sides are:

A = 15   B = 20   C = 25

D = 30   E = 40   F = 50

D = 3    E = 4     F = 5

6 0
2 years ago
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