Remark work with the two numbers that are 10^4 first. Then add in the one that goes to 10^3
Method one
You could change these all to integers and then take it back to scientific notation.
83 000
<u>-19 000</u>
64 000
<u>- 2 500
</u> 61 500 <u /><<<<<< answer
Method two
Keep the scientific notation.
8.3 * 10^4
<u>-1.9 * 10^4</u>
6.4 * 10^4
<u>- .25*10^4</u> Note the powers of ten must be the same.
6.15*10^4 <<<<<<<< answer
<u>
</u>Either method works.
<u>
</u>You could use your calculator to get the right answer. I'm not sure all calculators retain the scientific notation. Test yours to see what it does. Enter 8.3*10y^x 4 *1 =
<u>
</u>Your calculator might have x^y or ^ to record powers. You may also have EE and one of your keys.
<u>
</u>See if your calculator returns 8.3*10^4 or if it returns 83000<u>
</u>
Answer:
usual number of yellow eggs = 12
Usual maximum = 21
Usual minimum = 3
Step-by-step explanation:
To solve this, we will use the expected value of a binomial probability.
The formula is;
E(X) = np
Where;
n is sample size
p is probability of success.
We are given;
n = 58
p = 21%
Thus;
usual number of yellow eggs in samples = np = 58 × 21% = 12.18 ≈ 12
From USL(Upper specification limit) and LSL(Lower specification limit) formula, we can find the maximum usual number and minimum usual number of eggs respectively.
Thus;
USL = n(p + 3√(p(1 - p)/n)
USL = 58(0.21 + 3√(0.21(1 - 0.21)/58)
USL = 21.48 ≈ 21
LSL = n(p - 3√(p(1 - p)/n)
LSL = 58(0.21 - 3√(0.21(1 - 0.21)/58)
LSL = 2.87 ≈ 3
Mr. Sato needs 9 magnets.
Explanation:
AS there are 24 students in the science class,
there are <span><span>242</span>=12</span> pair of students.
Mr. Sato wants to give each pair of students 3 magnets
Mr. Sato has already given 9 pairs of students their 3 magnets
Therefore number of pairs of students yet to be given magnets is <span>12−9=3</span>.
For them he needs <span>3×3=9</span> magnets.
I think the mean with outliers excluded are called trimmean.
12/25 is 48% becuase 12/25 is alao 48/100 because 12 x 4= 48 and 25 x 4= 100.
