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3 years ago

6

Help come on plzz will give brainliest

1 answer:

BabaBlast [244]3 years ago

7 0

Answer:78.4

Step-by-step explanation:

(74+44+74+100+98+86+72+76+66+94)/10 = 78.4

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Length 21cm area 315cm2 find the breath

Ronch [10]

Answer:

Breadth = 15 cm

Step-by-step explanation:

Area = length x breadth

315 = 21 x breadth

$\frac{315}{21} = \frac{21}{21} \times breadth$ [ dividing both sides by 21 ]

$15 = 1 \times breadth\\\\breadth = 15 \ cm$

4 0

3 years ago

Read 2 more answers

Pls i need help haha

maksim [4K]

Answer:

i think the answer is 5

Step-by-step explanation:

6 0

3 years ago

You are shopping for a new pair of jeans and

kirill [66]

Answer:

7% percent off of $33.00 would be $30.69 cents

Hope helps

7 0

3 years ago

Find the slope of the line graphed in the diagram and choose the correct answer from the choices below. The two intercepts are (

kolezko [41]

1.2 is the slope of the line graphed in the given diagram

<u>Step-by-step explanation:</u>

We can use the given points (0, 3) and (-2.5, 0) to solve.

Slope formula is given by

$\text { slope, } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Given two intercepts:

$\left(x_{1}, y_{1}\right)=(0,3) \text { and }\left(x_{2}, y_{2}\right)=(-2.5,0)$

By substituting these in the slope equation, we get

$\text { slope, } m=\frac{0-3}{-2.5-0}=\frac{-3}{-2.5}=\frac{3}{2.5}=1.2$

5 0

4 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

4 years ago