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olga_2 [115]
3 years ago
8

What is the answer please

Mathematics
1 answer:
serg [7]3 years ago
4 0

Step-by-step explanation:

1a.\\ 8-y=2\qquad\text{subtract 8 from both sides}\\-y=-6\qquad\text{change the signs}\\\boxed{y=6}\\\\1b.\\11-9=2-b\\2=2-b\qquad\text{subtract 2 from both sides}\\0=-b\to \boxed{b=0}\\\\2a.\\5=\dfrac{c}{3}\qquad\text{multiply both sides by 3}\\15=c\to \boxed{c=15}\\\\2b.\\2=5-t\qquad\text{subtract 5 from both sides}\\-3=-t\qquad\text{change the signs}\\3=t\to\boxed{t=3}\\\\3a.\\5n=10\qquad\text{divide both sides by 5}\\\boxed{n=2}\\\\3b.\\a+6=12\qquad\text{subtract 6 from both sides}\\\boxed{a=6}

4a.\\v-10=7\qquad\text{add 10 to both sides}\\\boxed{v=17}\\\\4b.\\7=7n+7n\\7=14n\qquad\text{divide both sides by 14}\\\dfrac{7}{14}=n\\\boxed{n=\dfrac{1}{2}}\\\\5a.\\m+11=12\cdot4\\m+11=48\qquad\text{subtract 11 from both sides}\\\boxed{m=37}\\\\5b.\\6y=7+5\\6y=12\qquad\text{divide both sides by 6}\\\boxed{y=2}\\\\6a.\\b-8=3\qquad\text{add 8 to both sides}\\\boxed{b=11}\\\\6b.\\1+c=8\qquad\text{subtract 1 from both sides}\\\boxed{c=7}

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The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half its initial mass
Tju [1.3M]

Answer:

6.56 g

Step-by-step explanation:

We can write the mass of radioactive isotope left after a time t by using the equation

N(t)=N_0 (\frac{1}{2})^{\frac{t}{t_{1/2}}}

where

N_0 = 210 g is the initial amount of radioactive isotope

t_{1/2} is the half-life

t is the time

In this problem, we want to know the amount of radioactive isotope left after 5 half-lives, so after

t=5 t_{1/2}

Substituting into the equation, we have

N(t)=(210 g) (\frac{1}{2})^{\frac{5 t_{1/2}}{t_{1/2}}}=\frac{210 g}{2^5}=6.56 g

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3 years ago
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elena55 [62]
The answer to this is:

m = -13/3

Explanation:

Y2 - Y1 / X2 - X1

2-(-11) / -7-(-4)

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5 0
3 years ago
Read 2 more answers
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andrew11 [14]
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In a sample of 234 individuals over the age of 25, chosen at random from the state of Oregon, 48 did not have a high school dipl
Anna71 [15]

Answer:

The upper bound of a 95% confidence interval for the proportion of individuals over the age of 25 in Oregon who do not have a high school diploma is of 0.2568.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

95% confidence level

So \alpha = 0.05, z is the value of Z that has a p-value of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

In a sample of 234 individuals over the age of 25, chosen at random from the state of Oregon, 48 did not have a high school diploma.

This means that n = 234, \pi = \frac{48}{234} = 0.2051

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2051 + 1.96\sqrt{\frac{0.2051*0.7949}{234}} = 0.2568

The upper bound of a 95% confidence interval for the proportion of individuals over the age of 25 in Oregon who do not have a high school diploma is of 0.2568.

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