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3 years ago

What is 5x-15x + 20y = x + 5 Solve for y

Mathematics

2 answers:

rusak2 [61]3 years ago

5 0

$5x-15x +20y=x+5 \\ 
-10x+20y=x+5 \\ 
20y=11x+5 \\ 
y = \frac{11x+5}{20}$

Setler79 [48]3 years ago

4 0

Y=11x/20 +.25
Solve for y by simplifying both sides of the equation, then isolating the variable.

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Solve by using proper methods.

oksano4ka [1.4K]

Answer:

Approximately after 66.15 years, there will be 100 coyotes left

Step-by-step explanation:

We can use the formula $F=P(1+r)^t$ to solve this.

Where

F is the future amount (F=100 coyotes)

P is the initial amount (P=750 coyotes)

r is the rate of decrease per year (which is -3% per year or -0.03)

t is the time in years (which we need to find)

Putting all the information into the formula we solve.

Note: The logarithm formula we will use over here is $ln(a^b)=bln(a)$

So, we have:

$F=P(1+r)^t\\100=750(1-0.03)^t\\100=750(0.97)^t\\\frac{100}{750}=0.97^t\\\frac{2}{15}=0.97^t\\ln(\frac{2}{15})=ln(0.97^t)\\ln(\frac{2}{15})=tln(0.97)\\t=\frac{ln(\frac{2}{15})}{ln(0.97)}\\t=66.15$

Hence, after approximately 66.15 years, there will be 100 coyotes left.

Rounding, we will have 66 years

4 0

3 years ago

2x > -6 and x – 4 ≤ 3

Levart [38]

2x > -6

Divide both sides by 2.

x > -3

x - 4 ≤ 3

Add 4 to both sides.

x ≤ 7

6 0

3 years ago

A confidence interval (CI) is desired for the true average stray-load loss u (watts) for a certain type of induction motor when

Genrish500 [490]

Answer:

A) CI = (57.12 , 59.48)

B) CI = (57.71 , 58.89)

C) CI = (57.53 , 59.07)

D) n = 239.63

Step-by-step explanation:

given data:

mean, $\bar X = 58.3$

standard deviation, σ = 3

sample size, n = 25Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025,

Zc = Z(α/2) = 1.96

$ME = Zc * σ \sqrt{n}$

$ME = 1.96 * 3 \sqrt{25}$

ME = 1.18

$CI = (\bar X - Zc * s\sqrt{n} , \barX + Zc * s\sqrt{n})$

$CI = (58.3 - 1.96 * 3\sqrt{25} , 58.3 + 1.96 * 3\sqrt{25})$

CI = (57.12 , 59.48)

Given data:

mean, $\bar X = 58.3$

standard deviation, σ = 3

sample size, n = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

$ME = zc * σ \sqrt{n}$

$ME = 1.96 * 3\sqrt{100}$

ME = 0.59

$CI = (\bar X - Zc * s\sqrt{n} , \barX + Zc * s\sqrt{n})$

$CI = (58.3 - 1.96 * 3\sqrt{100} , 58.3 + 1.96 * 3\sqrt{100})$

CI = (57.71 , 58.89)

sample mean, $\bar X = 58.3$

sample standard deviation, σ = 3

sample size, n = 100

Given CI level is 99%, hence α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

$ME = Zc * σ \sqrt{n}$

$ME = 2.58 * 3\sqrt{100}$

ME = 0.77

$CI = (\bar X - Zc * s\sqrt{n} , \barX + Zc * s\sqrt{n})$

$CI = (58.3 - 2.58 * 3\sqrt{100} , 58.3 + 2.58 * 3/\sqrt{100}$

CI = (57.53 , 59.07)

Given data:

Significance Level, α = 0.01,

Margin or Error, E = 0.5,

σ = 3

The critical value for α = 0.01 is 2.58.

for calculating population mean we used

$n \geq (zc *σ/E)^2$

$n = (2.58 * 3/0.5)^2$

n = 239.63

7 0

3 years ago

Nancy shot a 24 on 4 holes of golf. At this rate, what can she expect her score to be if she plays all 18 holes?

Elden [556K]

Answer:

108

Step-by-step explanation:

"At this rate, ..." means we're to assume Nancy's score is proportional to the number of holes:

score/18 = 24/4

Multiplying by 18, we get ...

score = 18(24/4) = 18(6) = 108

Nancy can expect a score of 108 on 18 holes.

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3 years ago

A bag contains 15 marbles. The probability of randomly selecting a green marble is 1/5. The probability of randomly

meriva

Answer:

Step-by-step explanation:

well if you get all of the fractions into percentage form they will be 20%for green 8% for blue so in turn it would mean 8% of 15 is 2

3 0

3 years ago