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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?

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:

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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

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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