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nignag [31]
2 years ago
15

Expand (x-3)(x-8) to form a quadratic expression

Mathematics
1 answer:
AlladinOne [14]2 years ago
8 0

Answer:

x^2 - 11x + 24

Step-by-step explanation:

Foil the equation

x^2 - 3x - 8x + 24

x^2 - 11x + 24

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OK PLZ GIVE ME A GOOD ANSWER
AlekseyPX

Question:

At a dinner party, two deserts are being served. Six of the guests choose cheesecake, and eight of the guests choose apple pie. Write two associated part-to-whole ratios for this situation, in simplest form. Then, interpret these ratios within the situation.

Answer:

6 + 8 = 14

Prove:

You're diagrams going to need 14 pieces.

4 0
3 years ago
PLZ HELP PLZ 100 POINTS SO IMPORTANT BRAINLIEST
snow_tiger [21]

Answer:

only the first two i am pretty sure thw other ones make any sense  

Step-by-step explanation:

4 0
2 years ago
And English class has read the first 384 pages of your novel. If this is 75% of the entire book, how many total pages are in the
Doss [256]
384 pages (0.75) = 288 pages
Add this to the original amount of pages.

384 + 288 = 672 total pages.
3 0
2 years ago
Read 2 more answers
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
A student of the author earned grades of 92, 83, 77, 84, and 82 on herfive regular tests. She earned grades of 88 on the final e
goblinko [34]

The class is weighted as follows:

60% Regular Tests

10% Final Exam

15% Project

15% Homework

The total weighted points possible for the class are as follows:

(100+100+100+100+100)*.6 + 100*.1 + 100*.15 + 100*.15 = 340

To calculate her individual final weighted grade we plug in her scores for each category and complete the following computation:

(92+83+77+84+82)*.6 + 88*.1 + 95*.15 + 77*.15 = 285.4

So her weighted grade percent would be 285.4/340 = 83.9% which is a B.

8 0
3 years ago
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