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3 years ago

The statement "a is not more than b" is the same as a > b. True False

Mathematics

1 answer:

LenaWriter [7]3 years ago

3 0

Answer:

False

Step-by-step explanation:

A > B means A is greater than B, so A is not more than B is false.

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Find the value of x. The diagram is not drawn to scale. (1 point)

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cConsider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number,

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1. Kerry wants to paint her bedroom. The wall dimensions are 14 feet by 10 feet for

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Step-by-step explanation: what is it even asking?

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3 years ago

Solve the equation: 10 = -4 + x

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3 years ago

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4. The average annual income of 100 randomly chosen residents of Santa Cruz is $30,755 with a standard deviation of $20,450. a)

Umnica [9.8K]

Answer:

a) The standard deviation of the annual income σₓ = 2045

The calculated value Z = 0.608 < 1.645 at 10 % level of significance

Null hypothesis is accepted

The average annual income is greater than $32,000

The calculated value Z = 1.0977 < 1.96 at 5 % level of significance

Null hypothesis is accepted

The average annual income is equal to $33,000

d)

95% of confidence intervals of the Average annual income

(26 ,746.8 ,34, 763.2)

Step-by-step explanation:

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation = $20,450

The standard deviation of the annual income σₓ = $\frac{S.D}{\sqrt{n} }$

= $\frac{20,450}{\sqrt{100} }= 2045$

Given mean of the Population μ = $32,000

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation ( σ)= $20,450

Null Hypothesis:- H₀: μ > $32,000

Alternative Hypothesis:H₁: μ < $32,000

Level of significance α = 0.10

$Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{30755-32000 }{\frac{20450}{\sqrt{100} } }$

Z= |-0.608| = 0.608

The calculated value Z = 0.608 < 1.645 at 10 % level of significance

Null hypothesis is accepted

The average annual income is greater than $32,000

Given mean of the Population μ = $33,000

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation ( σ)= $20,450

Null Hypothesis:- H₀: μ = $33,000

Alternative Hypothesis:H₁: μ ≠ $33,000

Level of significance α = 0.05

$Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{30755-33000 }{\frac{20450}{\sqrt{100} } }$

Z = -1.0977

|Z|= |-1.0977| = 1.0977

The 95% of z -value = 1.96

The calculated value Z = 1.0977 < 1.96 at 5 % level of significance

Null hypothesis is accepted

The average annual income is equal to $33,000

d)

95% of confidence intervals is determined by

$(x^{-} - 1.96 \frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} })$

$(30755 - 1.96 \frac{20450}{\sqrt{100} } , 30755 +1.96 \frac{20450}{\sqrt{100} })$

( 30 755 - 4008.2 , 30 755 +4008.2)

95% of confidence intervals of the Average annual income

(26 ,746.8 ,34, 763.2)

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3 years ago