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3 years ago

Solve this expression and answer incorrect scientific notation (8.2E3)(4E5)

Mathematics

1 answer:

Anastaziya [24]3 years ago

6 0

Answer:

$(8.2E3)(4E5)=3.28E9$

Step-by-step explanation:

Given : Expression $(8.2E3)(4E5)$

To find : Solve this expression ?

Solution :

E denotes the power of 10,

$8.2E3=8.2\times 10^3$

$4E5=4\times 10^5$

Substitute in the expression,

$(8.2E3)(4E5)=(8.2\times 10^3)(4\times 10^5)$

$(8.2E3)(4E5)=(8.2\times 4)(10^3\times 10^5)$

$(8.2E3)(4E5)=(32.8)(10^8)$

$(8.2E3)(4E5)=(3.28\times 10^9)$

$(8.2E3)(4E5)=3.28E9$

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2.The number of tickets sold to a play for each showing is 78, 84, 87, 80, 91, 95, and 80.

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$Min =78$

The maximum is :

$Max = 95$

Now we can calculate the median since the sample size os n =7 the median would be the middle value at the position 4 from the dataset ordered and we got:

$Median =84$

Now we can find the quartile 1 we analyze the first 4 values 78, 80,80, 84 and the first quartile would be:

$Q_1= \frac{80+80}{2}= 80$

Now we can find the quartile 3 we analyze the last 4 values 84,87, 91, 95 and the third quartile would be:

$Q_3= \frac{87+91}{2}=89$

And finally the 5 number summary would be:

$Min = 78, Q_1 = 80, Median=84, Q_3 = 89, Max=96$

Step-by-step explanation:

We have the following data given:

78, 84, 87, 80, 91, 95, and 80.

The first step is order the dataset on increasing way and we got:

78, 80,80, 84,87, 91, 95

We can begin finding the minimum value and for this case is:

$Min =78$

The maximum is :

$Max = 95$

Now we can calculate the median since the sample size os n =7 the median would be the middle value at the position 4 from the dataset ordered and we got:

$Median =84$

Now we can find the quartile 1 we analyze the first 4 values 78, 80,80, 84 and the first quartile would be:

$Q_1= \frac{80+80}{2}= 80$

Now we can find the quartile 3 we analyze the last 4 values 84,87, 91, 95 and the third quartile would be:

$Q_3= \frac{87+91}{2}=89$

And finally the 5 number summary would be:

$Min = 78, Q_1 = 80, Median=84, Q_3 = 89, Max=96$

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3 years ago