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11Alexandr11 [23.1K]
2 years ago
8

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

is filled for each delivery, except for the last one, how many full truckloads are needed and how many cars will be in the last truck?
Mathematics
2 answers:
allochka39001 [22]2 years ago
5 0

divide 1006 by 12

1006 / 12 = 83.833

so 83 full trucks

83 *12 = 996

1006-996 = 10 cars will be on the last truck

Gennadij [26K]2 years ago
4 0

Answer:

Last one truckload will contain 10 cars and  total no. of truckloads required is 84

Step-by-step explanation:

Given :an auto transport truck holds 12 cars. A car dealer plans to bring in 1006 new cars in June and July.

To Find : . if an auto transport truck is filled for each delivery, except for the last one, how many full truckloads are needed and how many cars will be in the last truck?

Solution :

Let no. of truckloads used = x

Since each truckloads can hold 12 cars

So, x truckloads can hold 12x cars

since there are total 1006 cars

So, 12x=1006

x=\frac{1006}{12}

x = 83.83

So, No. of fully filled truckloads is 83

since 1 truckload holds 12 cars

So, 83 truckloads hold cars = 12*83 =996

Thus 996 cars are loaded in 83 truckloads

So, remaining cars = 1006-996 = 10 cars

So, one more truckload is required to load these remaining 10 cars.

So, last one truckload will contain 10 cars

So, total no. of truckloads requires = 83+1=84

Hence  last one truckload will contain 10 cars and  total no. of truckloads required is 84


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devlian [24]

Answer:

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Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

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In this problem, we have that:

A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm, so \mu = 49.9, \sigma = 3.74.

What is the length in cm of the longest 15% of Atlantic cod in this area?

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So

Z = \frac{X - \mu}{\sigma}

1.04 = \frac{X - 49.9}{3.74}

X - 49.9 = 3.8896

X = 53.7896

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abx - 5 = 0

Add 5 to both sides:
abx = 5

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x =  \frac{5}{ab}
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