Ask question

Ask question

All categories

3 years ago

13

Identify the percent of change as increase or decrease then find the percent of change round to the nearest tenth of a percent(

if necessary )
120 meals to 52 meals

1 answer:

harkovskaia [24]3 years ago

3 0

Decrease percentage = (120-52)/120 ×100 = 68/120 =56.66..%

On rounding off,

Percentage = 57%

HOPE THIS WILL HELP YOU

You might be interested in

Help me please! It’s so hard! Please answer!

Viktor [21]

Answer:

62

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Give another definition of variability and explain how it can be calculated

Kisachek [45]

${{\tt{••••••••••••••••••••••••••••••••••••••••••••••••••••}}}$

<h3 /><h3> DEFINITION OF<em> </em><em>VARIABILITY</em></h3>

» The degree to which data points in a statistical distribution or data collection deviate (variate) from the average value, as well as the degree to which these data points differ from one another.

$\blue{{\tt{ ⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩»⟩}}}$ ]

<h3>HOW CAN BE CALCULATED?</h3>

» Variability can be calculated by subtracting the lowest value from the highest value in the data set.

$\pink{{\tt{ ———————————————————————}}}$ ]

# $\cancel{CarryOnLearning}$

5 0

2 years ago

A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. As we sa

sashaice [31]

The question is incomplete so, here is the complete question.

A tank in form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. When friction and contraction of water at the hole are ignored, the height h of water in the tank is described by

$\frac{dh}{dt} = - \frac{A_{h} }{A_{w} } \sqrt{2gh}$ , where $A_{h}$ and $A_{w}$ are cross-sectional areas of the hole and the water, respectively. (a) Solve for h(t) if the initial height of the water is H. By hand, sketch the graph of h(t) and give its interval I of definition in terms of the symbols Aw, Ah and H. Use g = 32ft/s². (b) Suppose the tank is 12 feet high and has radius 4 feet and the circular hole har radius 1/2 inch. If the tank is initially full, how long will it take to empty?

Answer: a) h(t) = $(\sqrt{H} - 4 \frac{A_{h} }{A_{w} }t )^{2}$ , with interval 0≤t≤ $\frac{A_{w}\sqrt{H} }{4A_{h} }$

b) It takes 133 minutes.

Step-by-step explanation: a) The height per time is expressed as

$\frac{dh}{dt} = - \frac{A_{h} }{A_{w} } \sqrt{2gh}$

Using g=32ft/s²: $\sqrt{2gh}$ = $\sqrt{2.32.h}$ $= 8\sqrt{h}$

$\frac{dh}{dt} = - \frac{A_{h} }{A_{w} } \sqrt{2gh}$

$\frac{dh}{\sqrt{h} }$ = - 8. $\frac{A_{h} }{A_{w} }$ dt

$\int\limits^a_b {\frac{dh}{\sqrt{h} } } = - \int\limits^a_b {8.\frac{A_{h} }{A_{w} } } \, dt$

Calculating the indefinite integrals:

2 $\sqrt{h}$ = - 8 $\frac{A_{h} }{A_{w} }$ .t + c

According to the question, when t=0 h₀ for water is H. so

2 $\sqrt{H}$ = - 8 $\frac{A_{h} }{A_{w} }$ .0 + c

c = 2 $\sqrt{H}$

2 $\sqrt{h}$ = - 8 $\frac{A_{h} }{A_{w} }$ .t + 2 $\sqrt{H}$

Dividing each term by 2, then, the equation h(t) is

h(t) = ( $\sqrt{H} -$ $4\frac{A_{h} }{A_{w} } .t$ )²

Now, to find the interval, when the tank is empty, there is not water height so:

0 = ( $\sqrt{H} -$ $4\frac{A_{h} }{A_{w} } .t$ )²

0 = ( $\sqrt{H} -$ $4\frac{A_{h} }{A_{w} } .t$ )

4t $\frac{A_{h} }{A_{w} }$ = $\sqrt{H}$

t = $\frac{A_{w} \sqrt{H} }{4.A_{h} }$

Thus, the interval will be : 0 ≤ t ≤ $\frac{A_{w}\sqrt{H} }{4A_{h} }$

b) $r_{w}$ = 4 ft $h_{w}$ = 12 ft and $r_{h}$ = 1/2 in

Area of the water:

$A_{w} = \pi r^{2}$

$A_{w} = \pi .4^{2}$

$A_{w}$ = 16πft²

Area of the hole:

1 in = $\frac{1}{12}$ ft

so, $r_{h}$ = $\frac{1}{2}.\frac{1}{12}$ = $\frac{1}{24}$

$A_{h} = \pi .r^{2}$

$A_{h} = \pi .\frac{1}{24} ^{2}$

$A_{h}$ = $\frac{\pi }{576}$ ft²

As H=10 and having t = $\frac{A_{w} \sqrt{H} }{4.A_{h} }$ :

t = $\frac{16.\pi .\sqrt{12} }{4.\frac{\pi }{576} }$

t = 4.576. $\sqrt{12}$

t = 7981.3 seconds

t = $\frac{7981.3}{60}$ = 133 minutes

It takes 133 minutes to empty.

8 0

3 years ago

Triangle QRS is a right triangle with the right angle at vertex R. The sum of m∠Q and m∠S must be equal to 45°. greater than 45°

S_A_V [24]

According to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. If we know one angle is already 90, we do 180-90=90. The other 2 angles must add up to 90, exactly.

8 0

4 years ago

Read 2 more answers

Someone please help me

Gala2k [10]

The correct answer is, B. 1/5
Hope this helps!

7 0

3 years ago

Read 2 more answers