Lana is deciding whether to buy a pool pass for the summer. A pass costs $36.00 and

Average Price A stock analyst plots the price per share of a certain common stock as a function of time and finds that it can be

mixer [17]

Answer:

42.49 unit price

Step-by-step explanation:

The average price function, S(ave), will be an integral of the price function over the period of time (6 years), that is

S(ave) = $1/6*\int\limits^6_0 {S(t)} \, dt$

= $1/6*\int\limits^6_0 {37+6e^-0.03t} \, dt$

Solving the integral, we have

S(ave) = 1/6*(37t - (6e^-0.03t)/0.03) with the limits being from t = 0 to t = 6

Hence, we have

S(ave) = $37 - \frac{200}{6}*(e^(-0.03*6) - 1)$

This resolves to 42.49 unit price

8 0

3 years ago

Suppose a normally distributed set of stock prices with 3500 observations has a mean of 101 and a standard deviation of 10. Use

KengaRu [80]

Answer:

The number of observations in the data set expected to be between the values 91 and 121 is of 2853.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 101, Standard deviation = 10

Percentage of of observations in the data set expected to be between the values 91 and 121.

91 = 101 - 10

So 91 is one standard deviation below the mean.

121 = 101 + 2*10

So 121 is two standard deviations above the mean

The normal distribution is symmetric, which means that 50% of the measures are below the mean and 50% are above.

Of those 50% below the mean, 68% are between one standard deviation below the mean(91) and the mean(101).

Of those 50% above the mean, 95% are between the mean(101) and two standard deviations above the mean(121).

So the percentage of observations in this interval is of:

$P = 0.5*0.68 + 0.5*0.95 = 0.815$

Number of observations in the interval

81.5% of 3500. So

0.815*3500 = 2853

The number of observations in the data set expected to be between the values 91 and 121 is of 2853.

6 0

3 years ago

Rewrite this expression using a single exponent 10^2/10^5

andre [41]

Answer:

10^-3

Step-by-step explanation:

so if you have 10^2/10^5, you would subtract the exponents since it is division. 2-5=-3

therefore, it would be 10^-2

Hope this helps!

6 0

3 years ago

A high-speed computer printer prints a page in 1/6 second. Using this printer, how long would it take to print 30 pages?

german

$30*\frac{1}{6}=\frac{30}{6}=5$

6 0

4 years ago

Read 2 more answers

In the Midpoint Rule for triple integrals we use a triple Riemann sum to approximate a triple integral over a box B, where f(x,

Tomtit [17]

To approximate the volume with 8 boxes, we have to split up the interval of integration for each variable into 2 subintervals, [0, 1] and [1, 2]. Each box will have midpoint $m_{i,j,k}$ that is one of all the possible 3-tuples with coordinates either 1/2 or 3/2. That is, we're sampling $f(x,y,z)=\cos(xyz)$ at the 8 points,