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3 years ago

A study was done on several dogs that recently had litters of puppies. The results are shown below.

Mathematics

2 answers:

Reil [10]3 years ago

8 0

Answer: 1/5
Explanation: 8/40 simplified in half is 4/20 then 2/10 then 1/5!

AnnyKZ [126]3 years ago

6 0

8/40
8 divided by 8 = 1
40 divided by 8 = 5
so simplified, it would be 1/5

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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)

4 0

3 years ago

Please help will give brainlest

e-lub [12.9K]

Answer: ) (0,0), (0,2), (2,0), (4,2); Maximum value of 10

Hope this helps :)

7 0

3 years ago

Which of the following expressions could give the total amount of a $45 restaurant bill with a 15% tip added on?

aev [14]

$45 x .15 = $6.75 + $45 = $51.75

Hope this helps.

6 0

3 years ago

What are the coordinates of Image A (-2,7) after a dilation with a center at the origin and a scale factor of 3?

mr_godi [17]

Answer:

Step-by-step explanation:

7 0

3 years ago

I REALLY NEED HELP WITH THIS

luda_lava [24]

Answer:

x= 10

Step-by-step explanation:

Vertical angles theorem: 10x= 8x + 20

10x - 8x = 20 + 8x - 8x

2x = 20

x = 10

6 0

3 years ago