The area of D is given by:

$\int\limits \int\limits {1} \, dA = \int\limits_0^7 \int\limits_0^{x^2} {1} \, dydx \\ \\ = \int\limits^7_0 {x^2} \, dx =\left. \frac{x^3}{3} \right|_0^7= \frac{343}{3}$

The average value of f over D is given by:

$\frac{1}{ \frac{343}{3} } \int\limits^7_0 \int\limits^{x^2}_0 {4x\sin(y)} \, dydx = -\frac{3}{343} \int\limits^7_0 {4x\cos(x^2)} \, dx \\ \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49$

3 0

3 years ago

A cookie factory monitored the number of broken cookies per pack yesterday.

trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

$s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } }$ $\leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }$

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

8 0

3 years ago