HELP PLEASE <br> HOW DO I DO THIS?????<br> WHAT IS IT????

Once every three years, the Board of Governors of the Federal Reserve System collects data on household assets and liabilities t

kvv77 [185]

Answer:

C. The distribution of the value of transaction accounts is a right-skewed distribution.

Step-by-step explanation:

Some definitions

A skewed right distribution is one in which the tail is on the right side. On simple words we have most of the values on the left part of the distribution.

And present the following property:

$Mode< Median< Mean$

A skewed left distribution is one in which the tail is on the left side. On simple words we have most of the values on the left part of the distribution.

And present the following property:

$Mean< Median< Mode$

On a symmetric distribution we satisfy $Mean= MEdian= Mode$

For our case we have that $Mean= 270000$ and $Median=94500$

And we can see that $Mean> Median$ then we can conclude that this distribution is skewed to the right. And the best option would be:

C. The distribution of the value of transaction accounts is a right-skewed distribution.

6 0

3 years ago

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The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

6 0

3 years ago

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The question is in the picture.

wariber [46]

The answer is 37.....

7 0

3 years ago

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«<br> In simplified form, V18 is _

zysi [14]

Answer:

3V2

Step-by-step explanation:

V18

V9 times 2 equals 18

V3 to the second power because 3 times 3=9 so we get 3V2

6 0

3 years ago

During a football game, the visiting team was given 4 penalties for 15 yards each. What was the change in the visiting team’s fi

dlinn [17]

Answer:

60 yards back

Step-by-step explanation:Mutiply 4 and 15

5 0

3 years ago

HELP PLEASE HOW DO I DO THIS????? WHAT IS IT????