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maxonik
8 days ago
9

A compact disc is designed to last an average of 4 years with a standard deviation of 0.8 years. What is the probability that a

CD will last less than 3 years?
Mathematics
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Michael breeds chickens and ducks. Last month, he sold 505050 chickens and 303030 ducks for \$550$550dollar sign, 550. This mont
tester [12383]
Let the price of a chicken be represented as x and the price of a duck as y. The equations established are 50x + 30y = 550 (1) and 44x + 36y = 532 (2). By manipulating (1) with a factor of 6, we have 300x + 180y = 3,300 (3). Applying a factor of 5 to (2) results in 220x + 180y = 2,660 (4). Then, by subtracting (4) from (3), we find that 80x = 640, leading to x = 8. Substituting the value of x back into (1), we find 50(8) + 30y = 550, which simplifies to 30y = 150, and ultimately y = 5. Therefore, the costs are $8 for a chicken and $5 for a duck.
4 0
2 months ago
Consider 7 x 10 ^ 3 write a pattern to find the value of the expression
Leona [12618]
7 x 10^1 = 70
7 x 10^2 = 700
7 x 10^3 = 7000
3 0
2 months ago
Read 2 more answers
A set of angles has a ratio of 5 to 2. If the larger angle measures 105 degrees then what is the measure of the smaller angle
babunello [11817]

ratio 5:2

105, ___

5x = 105 → x = 21

2x = ____ → 2(21) = 42

Result: 42°

6 0
2 months ago
Read 2 more answers
ivy bought 10 packs of cups for her holiday party. a pack of medium cups costs $1.80 and a pack of large cups costs $2.40. she p
Leona [12618]
The answer is 4 packs of medium cups and 6 packs of large cups.
4 0
2 months ago
Read 2 more answers
10.4.1 .WP The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at ran
tester [12383]
Hello! You need to calculate a 99% confidence interval for the difference in mean lifespan between two tire brands. Each tested car was assigned one tire from each brand randomly on the rear wheels, allowing for paired sample analysis.       Brand 1  Brand 2  X₁-X₂      car 1:  36,925;  34,318;  2.607      car 2:  45,300;  42,280;  3.020      car 3:  36,240;  35,500;  0.740      car 4:  32,100;  31,950;  0.150      car 5:  37,210;  38,015;  -0.0805      car 6:  48,360;  47,800;  1.160      car 7:  38,200;  37,810;  0.390      car 8:  33,500;  33,215;  0.285      n= 8 The study variable is defined as Xd= X₁-X₂, where X₁ represents the tire lifespan (in km) from Brand 1 and X₂ represents Brand 2. Thus, Xd is the difference in tire lifespan. Xd~N(μd;δd²) (normality test p-value is 0.4640). For calculating the confidence interval, the best statistic is the Student's t using the following formula: t= (xd[bar] - μd)/(Sd/√n) ~t₍ₙ₋₁₎ sample mean: xd[bar]= 0.94 standard deviation: Sd= 1.29 = 3.355 xd[bar] ± t_{8;0.995}*(Sd/√n) ⇒ 0.94 ± 3.355*(1.29/√8) [-0.65;2.54]km. The CI can be compared to bilateral hypothesis testing: H₀:μd=0 H₁:μd≠0 using significance level of 0.01. Since the confidence interval includes zero, we do not reject the null hypothesis, indicating no significant difference between the tire brands. Hope you have a fantastic day!
5 0
2 months ago
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