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olga2289 [7]
3 years ago
7

Please help me with all of them. Plz plz and show me the work plz please

Mathematics
1 answer:
Natali [406]3 years ago
5 0
1:You have to multiply 1/2 × 12.You could do keep change flip so you will get 24 + 11 = 35

2: Repeat the same thing as I did with number one.
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The students from Mr. Jansen's class and MRs. Schmidt's class went on a trip toa science museum. Admission was $8 per student. T
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An airliner carries 50 passengers and has doors with a height of 70 in. Heights of men are normally distributed with a mean of 6
Korolek [52]

Answer:

dfdfdfdfdfd

Step-by-step explanation:

3 0
3 years ago
At a rate of 4% you paid $144 in interest over to year. What was the original amount of your loan
dem82 [27]
\bf ~~~~~~ \textit{Simple Interest Earned}
\\\\
I = Prt\qquad 
\begin{cases}
I=\textit{interest earned}\to &\$144\\
P=\textit{original amount deposited}\\
r=rate\to 4\%\to \frac{4}{100}\to &0.04\\
t=years\to &1
\end{cases}
\\\\\\
144=P(0.04)(1)\implies \cfrac{144}{(0.04)(1)}=P\implies 3600=P
8 0
3 years ago
A sample of a radioactive isotope had an initial mass of 360 mg in the year 1998 and
RUDIKE [14]

Answer:

193 mg

Step-by-step explanation:

Exponential decay formula:

  • A_t = A_0e^r^t
  • where Aₜ = mass at time t, A₀ = mass at time 0,  r = decay constant (rate), t = time  

Our known variables are:

  • 1998 to the year 2004 is a total of t = 6 years.
  • The sample of radioactive isotope has an initial mass of A₀ = 360 mg at time 0 and a mass of Aₜ = 270 mg at time t.

Let's solve for the decay constant of this sample.

  • 270=360e^-^r^(^6^)
  • 270=360e^-^6^r
  • \frac{3}{4} =e^-^6^r
  • \text{ln} (\frac{3}{4} )= \text{ln}(e^-^6^r)
  • \text{ln} (\frac{3}{4} )=-6r
  • r=-\frac{\text{ln}\frac{3}{4} }{6}
  • r=0.04794701

Using our new variables, we can now solve for Aₜ at t = 7 years, since we go from 2004 to 2011.

Our new initial mass is A₀ = 270 mg. We solved for the decay constant, r = 0.04794701.

  • A_t=270e^-^(^0^.^0^4^7^9^4^8^0^1^)^(^7^)
  • A_t=270e^-^0^.^3^3^5^6^2^9^0^7
  • A_t=193.01982213

The expected mass of the sample in the year 2011 would be 193 mg.

3 0
3 years ago
Given: ∆KLM, MF ⊥ KL MF = 15, KF = 12, FL = 19 Find: KM, ML, m∠K, m∠L, m∠KML
Svet_ta [14]

Answer:

Step-by-step explanation:

6 0
4 years ago
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