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FinnZ [79.3K]
2 years ago
11

Suppose that the halflife of a substance is 250 years . If there were ir 100 the substance , (a) give an exponential model for t

he situation, and (b) how much wil years?
can anyone answer this please:(
Mathematics
1 answer:
IgorLugansk [536]2 years ago
5 0
After 500 years one would have N nuclei of the original substance.

N = 100 * exp ( -ln(4) ) = 25

LOL.

I guess it is exp( - lamda * t)

Lamda, the decay constant is the ln(2) over the half life.

So we get exp( -ln(2)/250 * 500) or exp ( -ln(4) )
690 viewsView 1 Upvoter
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What is the distance between the points (-4,0) and (-7, -14) ?​
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Step-by-step explanation:

(-4,0) ; (-7, -14)

d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\\\ =\sqrt{(-7-[-4])^{2}+(-14-0)^{2}}\\\\ =\sqrt{(-7+4)^{2}+(-14)^{2}}\\\\=\sqrt{(-3)^{2}+(-14)^{2}}\\\\=\sqrt{9+196} \\\\=\sqrt{205} \\\\=14.3178

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5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim
Ne4ueva [31]

Answer:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

S_p=2.940

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

df=60+72-2=130

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

n_1 =60 represent the sample size for people who used a self service

n_2 =72 represent the sample size for people who used a cashier

\bar X_1 =5.2 represent the sample mean for people who used a self service

\bar X_2 =6.1 represent the sample mean people who used a cashier

s_1=3.1 represent the sample standard deviation for people who used a self service

s_2=2.8 represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

And t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

System of hypothesis

Null hypothesis: \mu_1 \geq \mu_2

Alternative hypothesis: \mu_1 < \mu_2

This system is equivalent to:

Null hypothesis: \mu_1 - \mu_2 \geq 0

Alternative hypothesis: \mu_1 -\mu_2 < 0

We can find the pooled variance:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

And the deviation would be just the square root of the variance:

S_p=2.940

The statistic is given by:

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

The degrees of freedom are given by:

df=60+72-2=130

And now we can calculate the p value with:

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

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Answer:

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Answer:

Step-by-step explanation:

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