How to mulitply 591 ×72

How dose using scientific Notion help compare when comparing with very small very large numbers

rewona [7]

Scientific notation is just a way to simplify a very large and oftentimes complicated number to a smaller one with the exact same value. For example, if I am given the number 1,656,000 to simplify using scientific notation, I would write it out as 1.656 x 10^6

The value at the end is exactly the same as the number I started out with. It is just written differently.

4 0

4 years ago

Factorise : (a+b)^2+4(a+b)+4

harina [27]

Answer:

2(3+2a+2b)

Step-by-step explanation:

6+4(a+b)

2(3+2(a+b))

4 0

3 years ago

The data below are the frequency of cremation burials found in 17 archaeological sites. a. Obtain the mean, median, and mode of

Sergeeva-Olga [200]

Answer:

(a) The mean of the data is 275.

(b) The median of the data is 85.

(c) The mode of the data is 45.

(d) The measure of center that works best here will be median.

Step-by-step explanation:

We are given below the frequency of cremation burials found in 17 archaeological sites. Arranging those in <u>ascending order</u> we get;

28, 31, 32, 45, 45, 47, 59, 67, 85, 86, 143, 256, 272, 390, 424, 524, 2141.

(a) The formula for calculating mean for the above data is given by;

Mean, $\bar X$ = $\frac{\sum X}{n}$

= $\frac{28+ 31+ 32+ 45+ 45+ 47+ 59+ 67+ 85+ 86+ 143+ 256+ 272+ 390+ 424+ 524+ 2141}{17}$

= $\frac{4675}{17}$ = 275

So, the mean of the data is 275.

(b) Since, the number of observations (n) in our data is odd (i.e. n = 17) , so the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

= $(\frac{17+1}{2} )^{th} \text{ obs.}$

= $(\frac{18}{2} )^{th} \text{ obs.}$

= $9^{th} \text{ obs.}$ = 85

So, the median of the data is 85.

(c) <u>Mode</u> is that value in our data which appears maximum number of times in our data.

So, after observing our data we can see that only number 45 is appearing maximum number of times (2 times) and all other numbers are appearing only once.

So, the mode of the data is 45.

(d) The measure of center that works best here will be median because there are outliers in our data (means extreme values) and mean gets affected by the outliers.

So, the best measure would be median as it represents the middle most value of our data.

4 0

3 years ago

Can someone pls explain I need help

erastovalidia [21]

Answer: 25

Step-by-step explanation: I just entered the equation into my calculator like so.

A negative number times a negative number equals a positive number.

I hope this helps, have a nice day.

4 0

3 years ago

Triphasil-28 birth control tablets are taken sequentially, 1 tablet per day for 28 days, with the tablets containing the followi

bazaltina [42]

Answer:

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

Step-by-step explanation:

The total milligrams of levonorgestrel is:

$L = L_{1} + L_{2} + L_{3}$

In which $L_{1}, L_{2}$ and $L_{3}$ are the number of miligrams of levonorgestrel taken in each phase.

The total milligrams of ethinyl estradiol is:

$E = E_{1} + E_{2} + E_{3}$

In which $E_{1},E_{2}$ and $E_{3}$ are the number of miligrams of ethinyl estradiol taken in each phase.

We can solve this by phase.

Phase 1: 6 tablets, each containing 0.050 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

$6*0.050mg = 0.30mg$ of levonorgestrel and $6*0.030mg = 0.18mg$ of ethinyl estradiol are taken.

So $L_{1} = 0.30$ and $E_{1} = 0.18$

Phase 2: 5 tablets, each containing 0.075 mg of levonorgestrel and 0.040 mg ethinyl estradiol

In this phase,

$5*0.075mg = 0.375mg$ of levonorgestrel and $5*0.040mg = 0.20mg$ of ethinyl estradiol are taken.

So $L_{2} = 0.375$ and $E_{2} = 0.20$

Phase 3: 10 tablets, each containing 0.125 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

$10*0.125mg = 1.25mg$ of levonorgestrel and $10*0.030mg = 0.30mg$ of ethinyl estradiol are taken.

So $L_{3} = 1.25$ and $E_{3} = 0.30$

The total milligrams of levonorgestrel is:

$L = L_{1} + L_{2} + L_{3} = 0.30 + 0.375 + 1.25 = 1.925$

The total milligrams of ethinyl estradiol is:

$E = E_{1} + E_{2} + E_{3} = 0.18 + 0.20 + 0.30 = 0.68$

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

6 0

3 years ago