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valentinak56 [21]
1 year ago
13

A substance used in biological and medical research is shipped by air freight to users in cartons of 1,000 ampules. The data bel

ow, involving 10 shipments, were collected on the number of times the carton was transferred from one aircraft to another over the shipment route (X) and the number of ampules found to be broken upon arrival (Y). Assume that first-order regression model (1.I) is appropriate. 2 5 7 8 9 10 Xi: Y: 16 9 17 12 22 13 8 15 19 11 0 2 0 2 2.15. Refer to Airfreight breakage Problem 1.21. a Because of changes in airline routes, shipments may have to be transferred more frequently than in the past. Estimate the mean breakage for the following numbers of transfers: X-2, 4. Use separate 99 percent confidence intervals. Interpret your results. b. The next shipment will entail two transfers. Obtain a 99 percent prediction interval for the number of broken ampules for this shipment. Interpret your prediction intervat c. In the next several days, three independent shipments will be made, each entailing two transfers. Obtain a 99 percent prediction interval for the mean number of ampules broken in the three shipments. Convert this interval into a 99 percent prediction interval for the total number of ampules broken in the three shipments.

Mathematics
1 answer:
givi [52]1 year ago
4 0

Answer:

Please see attachment

Step-by-step explanation:

Please see attachment

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Robert has $50 to spend on his utility bills each month. The basic monthly charge is 23.77 electricity costs 0.1117 for each kil
Lilit [14]

Answer:

<em>The maximum number of kilowatt-hours is 235</em>

Step-by-step explanation:

<u>Inequalities</u>

Robert's monthly utility budget is represented by the inequality:

0.1116x + 23.77 < 50

Where x is the number of kilowatts of electricity used.

We are required to find the maximum number of kilowatts-hours used without going over the monthly budget. Solve the above inequality:

0.1116x + 23.77 < 50

Subtracting 23.77:

0.1116x < 50 - 23.77

0.1116x < 26.23

Dividing by 0.1116:

x < 26.23/0.1116

x < 235

The maximum number of kilowatt-hours is 235

7 0
1 year ago
Which phrase best describes the translation from the graph y = 2(x – 15)2 + 3 to the graph of y = 2(x – 11)2 + 3? 4 units to the
Mama L [17]
Left is plus right is negative -11-(-15)=4  so you know 4 to the left
8 0
2 years ago
Read 2 more answers
What is the equation of the following line? Be sure to scroll down first to see all answer options. (0,3) (3,0) A. y = -x + 3 B.
qaws [65]
The correct answer is Choice A.

If you plot the points on a graph, you will see that there is a slope of -1 and the y-intercept is (0, 3).

This matches the equation of y = -x + 3 in Choice A.
4 0
2 years ago
The sum of two numbers is 0. Twice the smaller number subtracted from 3 times the larger number is 10. Let x represent th larger
slava [35]
The sum of two numbers is zero.
x + y = 0
y = -x

<span>Twice the smaller number subtracted from 3 times the larger number is 10.
Let x represent the larger number and y represent the smaller number.

Twice the smaller number: 2y

3 times the larger number: 3x

</span>Twice the smaller number subtracted from 3 times the larger number is 10. 
3x - 2y = 10

-2y = -3x + 10

y = 3/2 x - 5

The equations are:

y = -x
y = 3/2 x - 5

The answer is the first choice.
5 0
2 years ago
Read 2 more answers
Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean
Zinaida [17]

Answer:

E[R] = 99 Ω

\sigma_R = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, E[R] = \frac{a+b}{2}  = \frac{95+103}{2} = \frac{198}{2} = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333

Hence,

The standard deviation of resistance, \sigma_R = \sqrt{V(R)} = \sqrt5.333 = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

z_1 = \frac{102-99}{2.3094} = 1.299

z_2 = \frac{98-99}{2.3094} = -0.433

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

5 0
2 years ago
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