Solve for n. 11(n-1)+35=3n

Mathematics

1 answer:

kotykmax [81]3 years ago

4 0

Answer:

Move all terms to the left side and set equal to zero. Then set each factor equal to zero.Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

Solution

= − 3

11−11+35=3 11+24=3

11+24=311n+24=3n11n+24=3n11+24−24=3−24 1

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The Paralyzed Veterans of America is a philanthropic organization that relies on contributions. They send free mailing labels an

lions [1.4K]

Answer:

Confidence interval: (0.04649,0.04913)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 100,000

Number of people who donated, x = 4781

$\hat{p} = \dfrac{x}{n} = \dfrac{4781}{100000} = 0.04781
$

95% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting the values, we get:

$0.04781
\pm 1.96(\sqrt{\dfrac{0.04781
(1-0.04781
)}{100000}})\\\\=0.04781
\pm 0.00132\\\\=(0.04649,0.04913)$

is the required 95% confidence interval for the true proportion of their entire mailing list who may donate.

3 0

3 years ago

Please help fast......

Tom [10]

$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\ 1.AY=BX&\text{1. Given}\\ 2.AB \cong AB&\text{2. Reflexive Property}\\ 3. AD || BC&\text{3. Property of a square}\\ 4. \angle ABE \cong \angle AXB&\text{4. Alternate Interior Angles}\\ 5. \angle BAY \cong \angle BYA&\text{5. Alternate Interior Angles}\\6. \triangle BAX \cong \triangle ABY&\text{6. Angle-Side-Angle Theorem}\\ 7. AX \cong BY&\text{7. CPCTC}\\\end{array}$

*************************************************************************************

$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\1. AB=CF&\text{1. Given}\\2.AB+BF=A'F&\text{2. Segment Addition Postulate}\\3.CF+BF=A'F&\text{3. Substitution Property}\\4.CF+BF+BC&\text{4. Segment Addition Postulate}\\5.A'F=BC&\text{5. Transitive Property}\\6. \angle AFE = \angle DBC&\text{6. Given}\\7. EF = BD&\text{7. Given}\\8. \triangle AFE \cong \triangle CBD&\text{8. Side-Angle-Side Theorem}\\\end{array}$

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$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\\text{1.AC bisects }\angle BAD&\text{1. Given}\\2. \angle BAC \cong \angle DAC&\text{2. Property of angle bisector}\\3.AC = AC&\text{3. Reflexive Property}&4. \angle ACB \cong \angle ACD&\text{4. Property of angle bisector}\\5. \triangle ABC \cong \triangle ADC&\text{5. Angle-Side-Angle Theorem}\\6.BC=CD&\text{6. CPCTC}\\\end{array}$