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Tomtit [17]
3 years ago
8

Eric has recorded rainfall data for 90 days so far if this is 1/6 of the data needed for his science fair project for how many t

otal days will he record data
Mathematics
1 answer:
Mrrafil [7]3 years ago
3 0
Answer: 90•6=540
Explanation: If 1/6 of the data is 90 then we only have to multiply 90 with 6.
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A group of 3 adults and 5 children pay a total of $52 for movie tickets. A group of 2 adults and 4 children pay a total of $38 f
Ksivusya [100]
<span>x = child ticket price
y = adult ticket price 
5x+3y=52 the cost accounting for the first group.
3x+2y=38 the cost accounting for the second group.

Child price: 5$
Adult price: 9$</span>
3 0
3 years ago
Help me please I am begging y’all
cupoosta [38]
2 is the answer i think
7 0
2 years ago
A radioactive substance decays exponentially: The mass at time t is m(t) = m(0)ekt, where m(0) is the initial mass and k is a ne
velikii [3]

Answer:

M = 1/0.000121 = 8264.5 years

Step-by-step explanation:

M = − k ∫∞₀ teᵏᵗdt

To obtain this mean life, we'll use integration by parts to integrate the function ∫ teᵏᵗdt

∫udv = uv - ∫ vdu

u = t

du/dt = 1

du = dt

∫ dv = ∫ eᵏᵗdt

v = eᵏᵗ/k

∫udv = ∫ teᵏᵗdt

uv = teᵏᵗ/k

∫ vdu = eᵏᵗ/k

∫ teᵏᵗdt = (teᵏᵗ/k) - ∫eᵏᵗ/k

But, ∫eᵏᵗ/k = (1/k) ∫eᵏᵗ = (1/k²) eᵏᵗ = eᵏᵗ/k²

∫ teᵏᵗdt = (teᵏᵗ/k) - eᵏᵗ/k²

The rest of the calculation is done on paper in the image attached to this question

3 0
3 years ago
How many years will it take for the account to reach $18,600? Round your answer to the nearest hundredth
Margarita [4]
 <span>it depends how the interest is calculated, but there's not much of a difference 

assuming its continuously compouned, you use this formula: A(t)=Pe^(rt), where A is the final amount, P is the initial investment, r is the interest, and t is the time in years 

you want to find t such that A(t)=18,600 so 18,600=1000e^(.0675t) 

you need to use logarithm to figure it out, take the natural log of both sides 

the following properties will come into use: 

ln(a*b)=ln(a)+ln(b) 
ln(a^b)=bln(a) 
ln(e)=1 

taking the natural log 

ln(18,600)=ln(1000e^(.0675t)) 

ln(18,600)=ln(1000)+ln(e^.0675t) 

ln(18600)=ln(1000) + .0675t 

now solve for t: t= (ln(18600)-ln(1000))/.0675 

t=43.31</span>
4 0
2 years ago
How are the sun rays like mathematical rays
Ludmilka [50]
A ray in mathematics starts at a point and extends infinitely in one direction away from that point
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4 0
3 years ago
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