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Leviafan
1 month ago
13

You need to haul a load of patio bricks to a job site. each brick weighs 4 pounds 14 ounces. your truck can carry a 3/4-ton load

. how many bricks can your truck carry in a full load?
Mathematics
1 answer:
PIT_PIT [12.4K]1 month ago
6 0
To solve this problem, we need to convert the measurements so we can carry out the necessary calculations. A ton equals 2000 pounds, therefore, 3/4 ton translates to 1500 pounds. Since there are 16 ounces in a pound, 4 pounds and 14 ounces amounts to 4.875 pounds. To find the number of bricks, we divide:
                                     number of bricks = 1500 lbs / 4.875 lbs
                                                            = 307.7

Consequently, this is the number of bricks required. The final answer is approximately.
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an auto transport truck holds 12 cars. A car dealer plans to bring in 1006 new cars in June and July. if an auto transport truck
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Divide 1006 by 12

1006 / 12 = 83.833

This results in 83 complete trucks

83 * 12 = 996

Hence, 10 cars will be in the final truck

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Brian multiplied his average speed of 14.73 miles per hour by the amount of time he spent cross-country skiing. His product was
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Yes, it was a reasonable result

Step-by-step clarification:

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The probability that Jane will go to a ballgame (event A) on a Monday is 0.73, and the probability that Kate will go to a ballga
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Events A and B are termed independent when

Pr(A\cap B)=Pr(A)\cdot Pr(B),

if not, events A and B are classified as dependent.

The events A, B and A∩B are:

  • A - Jane plans to attend a ballgame on Monday;
  • B - Kate intends to go to a ballgame on Monday;
  • A∩B - Both Kate and Jane will be at the ballgame on Monday.

Pr(A)=0.73,\ Pr(B)=0.61,\ Pr(A\cap B)=0.52.\\ \\ Pr(A)\cdot Pr(B)=0.73\cdot 0.61=0.4453\neq 0.52=Pr(A\cap B).

Answer: events A and B are dependent


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1 month ago
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Axline Computers manufactures personal computers at two plants, one in Texas and the other in Hawaii. The Texas plant has 40 emp
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Answer:

a) The likelihood that none of the sampled employees are from the Hawaii plant is 1.74%.

b) The chance that exactly 1 employee from the sample is found working in the Hawaii plant is 8.70%.

c) There is an 89.56% chance that 2 or more employees in the sample are from the Hawaii plant.

d) The probability that 9 employees from the sample are working at the Texas plant is 8.70%.

Step-by-step explanation:

Each employee has two potential employment locations: either Texas or Hawaii. Thus, the binomial probability distribution can be utilized to solve this scenario.

Binomial probability distribution

This distribution defines the probability of achieving exactly x successes in n trials where there are only two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

Here, C_{n,x} denotes the number of ways to choose x objects from a set of n, represented by the subsequent formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of success occurring.

In this context, we know:

The sample comprises 10 employees, therefore n = 10.

a. Calculate the probability that none of the sampled employees are from the Hawaii plant (to 4 decimals)?

Given that 20 out of 60 employees are based in Hawaii:

p = \frac{20}{60} = 0.333

We aim to find P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.333)^{0}.(0.667)^{10} = 0.0174

Thus, the likelihood that none in the sample are from Hawaii stands at 1.74%.

b. Calculate the probability that 1 employee from the sample is from the Hawaii plant?

This is represented as P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{10,1}.(0.333)^{1}.(0.667)^{9} = 0.0870

Therefore, there is an 8.70% possibility that 1 employee in the sample comes from Hawaii.

c. Calculate the probability that 2 or more employees in the sample are from the Hawaii plant?

We can observe two scenarios: either fewer than 2 employees are from Hawaii or 2 and beyond. The combined probabilities equal decimal 1. So:

P(X < 2) + P(X \geq 2) = 1

We seek to find P(X \geq 2).

P(X \geq 2) = 1 - P(X < 2)

From problems a and b, we possess values for both probabilities.

P(X < 2) = P(X = 0) + P(X = 1) = 0.0174 + 0.0870 = 0.1044

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1044 = 0.8956

Accordingly, the chance that 2 or more employees in this sample operate at the Hawaii plant is 89.56%.

d. Calculate the likelihood that 9 employees in the sample are working at the Texas plant?

This corresponds to the probability found in part b for 1 employee working in Hawaii.

Consequently, there is an 8.70% chance that 9 employees belong to the Texas plant.

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