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3 years ago

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Six bags of soil are used to fill 5 flower pots. how much soil does each flower pot use? between what two whole numbers does the

answer lie?each flower pot uses ? or ? bags of soilso, 6 ÷ 5= ?the answer is between the whole numbers _1__ and _2__.

2 answers:

Zarrin [17]3 years ago

7 0

So Basically It's 5×6?

garri49 [273]3 years ago

7 0

Answer:

,,,,,,

Step-by-step explanation:

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A and B are discrete random variables. A can take on one of 20 possible values. B can take on one of 64 possible values. (In oth

deff fn [24]

Answer:

It defines 1280 possible outcomes

Step-by-step explanation:

A can take on one of 20 possible values.Then, B can take on one of 64 possible values

The form to choose these pairs of outcomes are

(Possible values from A) x (Possible values from B) = 20 x 64 = 1280

Also the sum of these 1280 pairs of values must be equal to 1 because P(A,B) it is a joint distribution of probability.

We sum this 1280 pairs of values because A and B are discrete random variables.Otherwise we would use integrals.

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3 years ago

Kerry has 224 inches of fabric to make shirts. She makes 5 identical shirts and has 16 inches of fabric remaining. How many inch

hodyreva [135]

224-16=208
208÷5=41.6
41.6 inches per shirt.

3 0

3 years ago

I'll give brainly pls help

Effectus [21]

Yes, option c is correct

6 0

3 years ago

Read 2 more answers

A map has a scale of 2 inches = 50 miles. If the

Inessa05 [86]

Answer:

175 miles apart.

Step-by-step explanation:

This is because 2 goes into seven 3 times and half of 50 is 25 so for 1 inch it would be 25 miles. 50+50+50+25=175 miles.

3 0

3 years ago

A mass weighing 13 lb stretches a spring 4.5 in. The mass is also attached to a damper with coefficient Y. Determine the value o

Gemiola [76]

Answer:55.227

Step-by-step explanation:

Given data

mass $\left ( m\right )$ =13lb

spring stretches by $\left ( x\right )$ =4.5in=0.375 ft

g= $32ft/s^2$

Now Spring constant K

mg=kx

k= $\frac{13\times 32}{y}=58.667lb/ft$

For critical damping $\zeta =1$

$2\times \zeta \times \omega_n =\frac{c}{m}$

and $\omega_n=\sqrt {\frac{k}{m}}$

$\omega _n=2.1241rad/s$

substituting values

$2\times 1 \times 2.124 =\frac{c}{m}$

c=55.227 lb.sec/ft

5 0

3 years ago