Ask question

Ask question

All categories

3 years ago

7

A radioactive substance decays exponentially with a half-life of 500 years. After 1500 years, what percent of the radioactive ma

terial remains?

1 answer:

Talja [164]3 years ago

7 0

12.5%, every 500 years you take away half of the substance (50%) -> 25% -> 12.5%
you divide your number 3 times because 1500/500=3

You might be interested in

BD is a diameter of circle Q. If M_ACB=30°, what is M<DBA?

Papessa [141]

I also agree with answer C

8 0

3 years ago

Which of the following expressions could give the total amount of a $45 restaurant bill with a 15% tip added on?

Sergio [31]

6.75 is the answer.........

8 0

3 years ago

You spend too much on your credit card and max it out. You have a credit limit of $5,000. The credit card has an annual APR of 1

vaieri [72.5K]

Answer:

APR credit card deal

Step-by-step explanation:

3 0

3 years ago

A machine produces a part for the automotive industry. 4% of the parts produced were defective in the past, and we believe that

Mamont248 [21]

Answer:

$n=\frac{0.04(1-0.04)}{(\frac{0.03}{1.64})^2}=114.76$

And rounded up we have that n=115

Step-by-step explanation:

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

Solution to the problem

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

The proportion of defectives is estimated as: $\hat p=0.04$ . And on this case we have that the margin of error is $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.04(1-0.04)}{(\frac{0.03}{1.64})^2}=114.76$

And rounded up we have that n=115

5 0

3 years ago

Evaluate. 2^−2⋅(12⋅3)−5^3

Oduvanchick [21]

$\text{Hello there!}\\\\\text{Solve the equation:}\\\\2^{-2}\left(12\cdot 3\right)-5^3\\\\0.25(12\cdot3)-5^3\\\\0.25(36)-5^3\\\\9-5^3\\\\9-125\\\\\large\boxed{-116}$

4 0

3 years ago

Read 2 more answers