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3 years ago

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A set of data whose histogram is extremely skewed yields a sample mean and standard deviation of 69.5 and 10.75, respectively. W

hat is the minimum percentage of observations that:
A. are between 48 and 91.

and

B. are between 37.25 and 101.75

1 answer:

frosja888 [35]3 years ago

3 0

I am a beautiful person who can help you with the new one of those who are interested

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1. Find the perimeter.<br> 13 yd<br> 14 yd<br> 7 yd<br> 18 yd

nikdorinn [45]

Answer:

52 yards

Step-by-step explanation:

Perimeter is all sides added together.

13 + 14 + 7 + 18 = 52

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2 years ago

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Shen’s mother gave him $10 to go buy groceries. He picked up a gallon of milk for $3.49, three pounds of oranges that cost $1.14

svet-max [94.6K]

Gallon of milk: $3.49
3lbs oranges: $1.14(3) = $3.42
Box of cereal: $3.46

Front end estimation is taking the number in front and the rest of the numbers coming after it will be zeroes.

Using front end estimation:
$3.49 => $3.00
$3.42 => $3.00
$3.46 => $3.00

Add all these up; Shen estimated the total cost of the groceries would be $9.00.

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AYUDAAAA, LES DOY CINCO ESTRELLAS SI LO RESUELVEN BIEN

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dividido por la epotenusa es 7 :v

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2 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

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3 years ago

Please solve for me and give me a caculator where i can find out how to do this

tekilochka [14]

We can help you and somebody can help you with calculator

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3 years ago