Answer:
523.59 or 523.60
Step-by-step explanation:
Volume for a sphere: 4/3πr^3 (4÷3x π x r ^ 3)
(r is radius, V is volume)
V= 4/3 x π x r^3
The diameter is 10, so the radius is 5...
V= 4/3 x π x 5^3
V= 4/3 x π x 125
CALCULATER: 4 ÷ 3 x π x 125
Answer 523.59 (523.60 rounded)
Answer:
Step-by-step explanation:
We proceed to show the procedure to calculate the given fraction into a decimal form:
1) Since numerator is less than denominator, the integer component of the decimal number is zero:
2) We multiply the numerator by 10 and find the tenth digit:
Then,
3) We multiply the fraction in 2) by 10 and find the hundredth digit:
Then,
And the remainder is:
4) We multiply the remainder by 10 and divide this result by the denominator to determine the thousandth digit:
Then,
This question asks us to write a decimal correct to 2 decimal places, which has the characteristic that is infinite periodical decimal. Then, the result correct to 2 decimal places is:
Answer:
x = 3.8
Step-by-step explanation:
The equation is (5 - x) / 2 = 2x - 7
Multiply both sides by 2:
5 - x = 4x - 14
Add 14 to both sides:
19 - x = 4x
Add x to both sides:
19 = 5x
Divide both sides by 5 to isolate x:
3.8 = x
Therefore, x = 3.8.
This question is incomplete, the complete question is;
Let X denote the time in minutes (rounded to the nearest half minute) for a blood sample to be taken. The probability mass function for X is:
x 0 0.5 1 1.5 2 2.5
f(x) 0.1 0.2 0.3 0.2 0.1 0.1
determine;
a) P( X < 2.5 )
B) P( 0.75 < X ≤ 1.5 )
Answer:
a) P( X < 2.5 ) = 0.9
b) P( 0.75 < X ≤ 1.5 ) = 0.5
Step-by-step explanation:
Given the data in the question;
The probability mass function for X is:
x 0 0.5 1 1.5 2 2.5
f(x) 0.1 0.2 0.3 0.2 0.1 0.1
a) P( X < 2.5 )
P( X < 2.5 ) = p[ x = 0 ] + p[ x = 0.5 ] + p[ x = 1 ] + p[ x = 1.5 ] + p[ x = 2 ]
so
P( X < 2.5 ) = 0.1 + 0.2 + 0.3 + 0.2 + 0.1
P( X < 2.5 ) = 0.9
b) P( 0.75 < X ≤ 1.5 )
P( 0.75 < X ≤ 1.5 ) = p[ x = 1 ] + p[ x = 1.5 ]
so
P( 0.75 < X ≤ 1.5 ) = 0.3 + 0.2
P( 0.75 < X ≤ 1.5 ) = 0.5
Answer:
Therefore we can say that 98 boxes will be needed to hold all the paperweights.
Step-by-step explanation:
i) there are 1657 souvenir paperweights that need to be packed in boxes.
ii) each box will hold 17 paperweights
iii) therefore the number of boxes that will be needed are
= 
= 
= 97.471
iv) Therefore we can say that 98 boxes will be needed to hold all the paperweights.
