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4 years ago

5

There are 4 semi-trailer trucks parked in a line at the rest stop. After the first truck, each truck in the line weighs 2 tons m

ore than the truck before it. The trucks weigh a total of 32 tons. How many pounds does each truck weigh?

2 answers:

butalik [34]4 years ago

4 0

Let the weight of the first truck be represented by w.

total weight = w + (2 + w) + (2 + 2 + w) + (2 + 2 + 2 + w)
32 = 4w + 12
20 = 4w
w = 5

Therefore, the first truck weighs 5 tons, the second weighs 7 tons, the third weighs 9 tons and the fourth weighs 11 tons.

3 years ago

boooooo

Xelga [282]4 years ago

3 0

Answer:

The weights of the trucks are 5, 7, 9 and 11 tons respectively.

Step-by-step explanation:

Let the weight of first truck = x ( in tons ).

It is given that the weight of the trucks is increasing by 2 tons after the first truck.

So, we get,

Truck Weight

1 x

2 x+2

3 x+2+2 = x+4

4 x+4+2 = x+6

Since, the total weights of the trucks is 32 tons. We get,

$x+(x+2)+(x+4)+(x+6)=32$

i.e. $4x+12=32$

i.e. $4x=20$

i.e. x= 5

So, the weight of the first truck is 5 tons.

Then, we have,

The weight of 2nd truck = 5+2 = 7 tons

The weight of 2nd truck = 5+4 = 9 tons

The weight of 2nd truck = 5+6 = 11 tons

Hence, the weights of the trucks are 5, 7, 9 and 11 tons respectively.

3 years ago

boooo stupid noon

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Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0

3 years ago

Suppose a certain computer virus can enter a system through an email or through a webpage. There is a 40% chance of receiving th

DedPeter [7]

Answer:

P = 0.42

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, that is the probability of receiving this virus through the email.

-Set B, that is the probability of receiving it through the webpage.

The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that $A \cap B$ = 0.17.

By email only

The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:

$A + (A \cap B) = 0.40$

$A + 0.17 = 0.40$

$A = 0.23$

where A is the probability that the system receives the virus just through the email.

The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.

By webpage only

There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:

$B + (A \cap B) = 0.35$

$B + 0.17 = 0.35$

$B = 0.18$

where B is the probability that the system receives the virus just through the webpage.

The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.

What is the probability that the virus does not enter the system at all?

So, we have the following probabilities.

- The virus does not enter the system: P

- The virus enters the system just by email: 23% = 0.23

- The virus enters the system just by webpage: 18% = 0.18

- The virus enters the system both by email and by the webpage: 17% = 0.17.

The sum of the probabilities is 100% = 1. So:

P + 0.23 + 0.18 + 0.17 = 1

P = 1 - 0.58

P = 0.42

There is a probability of 42% that the virus does not enter the system at all.

5 0

3 years ago

write the equation in slope intercept form of the line that passes through -1, 11 and is parallel to the graph of y = -8x + 3

lawyer [7]

<h2>Answer: The line from the question [ y = -8x + 3 ] passes through the point ( -1, 11 ). </h2>

<h3 /><h3>Step-by-step explanation: </h3>

<u>Find the slope of the parallel line</u>

When two lines are parallel, they have the same slope.

⇒ if the slope of this line = - 8

then the slope of the parallel line (m) = - 8

<u>Determine the equation</u>

We can now use the point-slope form (y - y₁) = m(x - x₁)) to write the equation for this line:

⇒ y - 11 = - 8 (x - (-1))

∴ y - 11 = - 8 (x + 1)

We can also write the equation in the slope-intercept form by making y the subject of the equation and expanding the bracket to simplify:

since y - 11 = - 8 (x + 1)

y = - 8 x + 3

The line from the question [ y = -8x + 3 ] passes through the point ( -1, 11 ).

5 0

3 years ago

Ind the slope and y intercept of -9x+3y=5

Radda [10]

Answer:

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4 0

3 years ago

I need help please and thank you

azamat

Answer:

I believe it's B ............

8 0

3 years ago