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lozanna [386]
3 years ago
9

Nevermind, the question has been solved.

Mathematics
2 answers:
san4es73 [151]3 years ago
7 0

okay.... I guess??? lol

Oksi-84 [34.3K]3 years ago
3 0

Answer:

Step-by-step explanation:

than im getting points ;)

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Hello there!
jok3333 [9.3K]

Answer:

4. dy/dx = -2

8. dy/dx = 1/2 x^(-3/2)

10/ dy/dr = 4 pi r^2

Step-by-step explanation:

4.  y = -2x+7

dy/dx = -2(1)

dy/dx = -2

8.  y = 4 - x^-1/2

 dy/dx =  - (-1/2x^ (-1/2-1)

 dy/dx = 1/2 x^(-3/2)

10.  y = 4/3 pi r^3

dy/dr = 4/3 pi  (3r^2)

dy/dr = 4 pi r^2

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3 years ago
Kerry is planning to plant a rectangular garden that is 16 feet longer than it wide. Kerry wants to find its width if its area i
trapecia [35]
I think it is A. I hope Im correct.
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3 years ago
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Just how much is a million?
Fudgin [204]
Hey You!

1 million = 1,000,000

In other words, 1 million is 1000 1000s.
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3 years ago
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You plan to borrow $36,500 at a 7.7% annual interest rate. The terms require you to amortize the loan with 7 equal end-of-year p
erma4kov [3.2K]

The amount of interest you would be paying in Year 2 is: $2,492.62.

<h3>Interest</h3>

First step is to calculate the Equal Monthly Payment

Equal Monthly Payment=P×r×(1+r)^t/(1+r)^t-1

Where:

P=Principal=$36,500

r=Rate=7.7%

t=Time=7 years

Equal Monthly Payment=36,500×0.077×(1+0.077)^7÷(1+0.077)^7-1

Equal Monthly Payment=36,500×0.077×(1.077)^7÷(1.077)^7-1

Equal Monthly Payment=36,500×0.077×1.6807763÷1.6807763-1

Equal Monthly Payment=4,723.82/0.6807763

Equal Monthly Payment=$6,938.875

Second step is to calculate Year 1 Closing balance

Year 1 Closing balance  = Beginning  balance + Interest - EMI Payment

Year 1 Closing balance=  $36,500 +($36,500×7.7%) - $6,938.875

Year 1 Closing balance=  $36,500 + $2,810.5 -$6,938.875

Year 1 Closing balance =   $32,371.625

Third step is to calculate year 2 interest

Year 2 Interest= $32,371.625×7.7%

Year 2 Interest=$2,492.62

Therefore the amount of interest you would be paying in Year 2 is: $2,492.62.

Learn more about interest here:brainly.com/question/15259578

#SPJ1

8 0
2 years ago
A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi
SVETLANKA909090 [29]

Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

= \frac{20!}{(20-5)! 5!)}

= \frac{20*19*18*17*16}{5*4*3*2*1}

= 15504

b) probabilities of A2, A5, A4

P(A2) = ⁵C₂ / ²⁰C₂

= \frac{20}{320} = \frac{1}{19}

P(A5) = ⁸C₅ / ²⁰C₅

= \frac{56}{15540} = \frac{7}{1938}

P(A4) = ⁷C₄ / ²⁰C₄

= \frac{35}{4845} = \frac{7}{969}

c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1

5 0
3 years ago
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