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3 years ago

Can someone pls help me with this its sooo confusing

Mathematics

1 answer:

Mademuasel [1]3 years ago

4 0

9514 1404 393

Answer:

$189.32

Step-by-step explanation:

Add all of the numbers shown to the beginning balance to find the ending balance.

$92.19 -39.35 +49.76 +87.35 +9.97 -10.60 = $189.32

Tristan's balance at the end of the month was $189.32.

There should be no confusion. Money deposited into the account increases the balance by the amount deposited.

Money spent from the account decreases the balance by the amount spent.

Here, amounts deposited are given a plus sign (+); amounts spent are given a minus sign (-).

Additional comment

It can be tedious to enter these numbers correctly into a calculator. Many folks prefer a calculator that provides a "tape", a record of the amounts and operations the calculator uses in the computation. Alternatively, you can put the numbers in a spreadsheet and use the SUM function to do the addition for you. That, too, makes it easy to check that numbers have been entered correctly.

The attachment shows the calculation done in a spreadsheet.

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FACTOR and list your 4 steps: 5x^2y - 30xy^2

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Answer:

5x • (x + 3y) • (x - 2y)

Step-by-step explanation:

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2 years ago

What is the value of t=1 ∑³ (4 x 1/2^t-1)

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Answer:

Step-by-step explanation:

$\sum\limits_{t=1}^{3}(4\cdot(\frac{1}{2})^{t-1})$

This is the sum of the first three terms of a geometric sequence, where the first term is 4 and the common ratio is ½.

We can use a formula to find the sum, or, since there's only three terms, we can find the value of each term then add up the results.

4 · (½)¹⁻¹ = 4

4 · (½)²⁻¹ = 2

4 · (½)³⁻¹ = 1

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3 years ago

16. How many different words can be formed with the letters of the word 'RAJARAM'? In how many of

Mandarinka [93]

Answer:

The word "ARRANGE" can be arranged in

2!×2!

5040

=1260 ways.

For the two R's do occur together, let us make a group of R's taking from "ARRANGE" and permute them.

Then the number of ways =

=360.

The number ways to arrange "ARRANGE", where two "R's" will not occur together is =1260−360=900.

Also in the same way, the number of ways where two "A's" are together is 360.

The number of ways where two "A's" and two "R's" are together is 5!=120.

The number of ways where neither two "A's" nor two "R's" are together is =1260−(360+360)+120=660.

Step-by-step explanation:

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3 years ago

Read 2 more answers

Alma was given some money as a gift and deposited it into a savings account. She makes a withdrawal of the same amount at the en

Kryger [21]

Answer:

input= starting amount

output= withdrawals

(5,442) and (13,226)

Step-by-step explanation:

for input its what you put in the account for output its for what you take out.

And for the two points it the months first then you put how much money was left .

Hope this helps!

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2 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