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MAVERICK [17]
3 years ago
8

A construction worker needs 3 2/3 pounds of concrete mix. He has 1 1/2 pounds of concrete mix in one bag, and he has 3 1/0 pound

of concrete mix in another bag.
How many more pounds of concrete mix does the construction worker need?

Enter your answer as a mixed number in simplest form by filling in the boxes.
Mathematics
1 answer:
Anvisha [2.4K]3 years ago
7 0
The answer is 4/5 because that’s how you solve the equation of life
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Please help me it's due today:(​
kvasek [131]
Hey man this is very simple please use photo math or search the answers up thanks man glad I could help in any way
4 0
2 years ago
The graph of F(x) shown below resembles the graph of G(x) = x ^ 2 but it has been changed somewhat . Which of the following coul
sukhopar [10]

Answer:

f(x) = 3x^{2} + 2

Step-by-step explanation:

Given G(x) = x^{2}

As we can see, the graph of f(x) is 2 units above the graphof g(x) in vertical (vertical shift)

=> f(x) = ax^{2} + 2

As you can see, the graph of f(x) is stretched vertically by a dilation factor of 3

and there is no reflection, so a= 3

=> the  equation of f(x) is:  f(x) = 3x^{2} + 2

5 0
3 years ago
At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a
morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = \dfrac{15}{120}=0.125

As we know the poisson process, we get that

P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda

So, for exactly one car would be

P(n=1) is given by

=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

0.2599\times 18=4.6798

We will find the traffic flow q such that

P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%

Hence, a) 4.6798, and b) 19.8%.

7 0
3 years ago
Find the value of x to the nearest tenth!
sergij07 [2.7K]

Answer:

5.7

Step-by-step explanation:

sine cosine tangent

soh cah toa

sine = opposite/hypotenuse

so sin(35)= x/10

you can multiply 10 to both sides to get rid of the 10 denominator on the right leaving you with 10sin(35)=x

be sure your calculator is in degrees.

put that into a calculator leaving you with 5.7

5 0
3 years ago
Which of the following expressions is equivalent to 3 m ( m − 2 ) − ( m ^2 + 1 ) ?
leonid [27]

Answer:

uh

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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