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2 years ago

10

A swimming pool and deck are built in a rectangle with length 40 ft and width 18 ft. If the pool itself is 320 square feet, what

is the area of the deck itself?

1 answer:

stich3 [128]2 years ago

8 0

320x2=640
64u0x18=435
ugddx bb v

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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)

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b)

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

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

<em>Null hypothesis is accepted </em>

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<em>d) </em>

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<em>(26 ,746.8 ,34, 763.2)</em>

<u>Step-by-step explanation:</u>

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation = $20,450

a)

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

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

b)

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

<u><em>Null Hypothesis:- H₀</em></u>: μ > $32,000

<u><em>Alternative Hypothesis</em></u>: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

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

<em>Null hypothesis is accepted </em>

<em>The average annual income is greater than $32,000</em>

c)

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

<u><em>Null Hypothesis:- H₀</em></u>: μ = $33,000

<u><em>Alternative Hypothesis</em></u>: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

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

<em>Null hypothesis is accepted </em>

<em>The average annual income is equal to $33,000</em>

<em>d) </em>

<em>95% of confidence intervals is determined by</em>

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

<em></em> $(30755 - 1.96 \frac{20450}{\sqrt{100} } , 30755 +1.96 \frac{20450}{\sqrt{100} })$ <em></em>

<em>( 30 755 - 4008.2 , 30 755 +4008.2)</em>

<em>95% of confidence intervals of the Average annual income</em>

<em>(26 ,746.8 ,34, 763.2)</em>

<em></em>

<em></em>

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2 years ago

Please help me answer this question

Illusion [34]

To determine if the ordered pair (3 , -2) is a solution to 4x = 3y + 18,

plug in the point into the equation.

4x = 3y + 18 (Plug in 3 for "x", and plug in -2 for "y")

4(3) = 3(-2) + 18

12 = -6 + 18

12 = 12

This is a solution because in the end, the number is the same on both sides

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2 years ago