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ycow [4]
3 years ago
11

A study was designed to test whether applying metal tags is detrimental to a penguin. One variable examined is the survival rate

10 years after tagging. The scientists observed that 10 of the 50 metal tagged penguins survived, compared to 18 of the 50 electronic tagged penguins. Construct a confidence interval for the difference in proportion surviving between the metal and electronic tagged penguins (). Let be the proportion of penguins with metal tags that survived and be the proportion of penguins with electronic tags that survived. Round your answers to three decimal places.
Mathematics
1 answer:
kicyunya [14]3 years ago
4 0

Complete question :

A study was designed to test whether applying metal tags is detrimental to a penguin. One variable examined is the survival rate 10 years after tagging. The scientists observed that 10 of the 50 metal tagged penguins survived, compared to 18 of the 50 electronic tagged penguins. Construct a 90% confidence interval for the difference in proportion surviving between the metal and electronic tagged penguins (). Let be the proportion of penguins with metal tags that survived and be the proportion of penguins with electronic tags that survived. Round your answers to three decimal places.

Answer:

(-0.305 ; - 0.015)

Step-by-step explanation:

Sample size, n1 = 50

x1 = 10

n2 = 50

x2 = 18

P1 = x1 / n1 = 10/50 = 0.2

P2 = x2 / n2 = 18 /50 = 0.36

1 - P1 = 0.8 ; 1 - P2 = 0.64

Zcritical at 90% = 1.645

Confidence interval :

(P1 - P2) ± Zcritical * standard error

Standard Error = sqrt[(p1(1-p1))/n1 + (p2(1-p2))/n2]

S.E = sqrt((0.2(0.8))/50 + 0.36(0.64))/50)

S.E = sqrt((0.0032 + 0.004608))

S. E = sqrt(0.007808)

S. E = 0.0883628

(0.2 - 0.36) ± (1.645 * 0.0883628)

-0.16 ± 0.145356806

Lower boundary = - 0.16 - 0.145356806 = −0.305356806

Upper boundary = - 0.16 + 0.145356806 = −0.014643194

(-0.305 ; - 0.015)

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torisob [31]

Answer:

Part A) 26\ liters of petrol and 4\ liters of oil

part B) 3.25\ liters or 3,250\ ml of petrol

Step-by-step explanation:

Part A) How much of each is there in 30 liters of mixture ?

Let

x-----> the amount of liters of petrol

y----> the amount of liters of oil

we know that

\frac{x}{y}=\frac{13}{2}

x=\frac{13}{2}y -----> equation A

x+y=30 ---> equation B

substitute equation A in equation B

\frac{13}{2}y+y=30

\frac{15}{2}y=30

y=30*2/15=4\ liters

Find the value of x

x=\frac{13}{2}(4)=26\ liters

Part B) How much petrol would be mixed with 500 ml of oil?

Convert ml to l

500\ ml=0.50\ l

we know that

using proportion

\frac{13}{2}=\frac{x}{0.5}\\ \\x=13*0.5/2\\ \\x=3.25\ l

3 0
3 years ago
Solve systems of equations for y and x: <br><br> −5y + 8x = −18<br><br> 5y + 2x = 58​
Natali5045456 [20]

Answer:

(4, 10)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

Equality Properties<u> </u>

  • Multiplication Property of Equality
  • Division Property of Equality
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  • Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

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Step-by-step explanation:

<u>Step 1: Define Systems</u>

-5y + 8x = -18

5y + 2x = 58

<u>Step 2: Solve for </u><em><u>x</u></em>

<em>Elimination</em>

  1. Combine 2 equations:                                                                                      10x = 40
  2. [Division Property of Equality] Divide 10 on both sides:                                x = 4

<u>Step 3: Solve for </u><em><u>y</u></em>

  1. Substitute in <em>x</em> [Original equation]:                                                                  -5y + 8(4) = -18
  2. Multiply:                                                                                                             -5y + 32 = -18
  3. [Subtraction Property of Equality] Subtract 32 on both sides:                      -5y = -50
  4. [Division Property of Equality] Divide -5 on both sides:                                y = 10
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