Y-3=-2(x+5) A.-13 B.-7 C.2 D.8

Simplify: 9(1 - r) + 3r

Butoxors [25]

Answer:

hdudhryhrgrhdidbf7hs6he8hs6jw9wg5gssi

4 0

3 years ago

Read 2 more answers

Julie drew a picture of her rectangular garden,. In her picture,the length of her garden is 7/8 inches and the width is 3/4 inch

stepan [7]

Answer:

21/32 square inches.

Step-by-step explanation:

The area of a rectangle = length x width

The area of her picture = length x widht

= 7/8 x 3/4

= 21/32

The area of her picture is 21/32 square inches.

Yes, I agree the area of the picture is 21/32 square inches.

The area must be in square inches. So she need to specify it.

Thank you.

5 0

3 years ago

Which is the best unit to measure the distance between califronia and new york

Ludmilka [50]

Answer:

Miles would be the best.

Step-by-step explanation:

Miles are the largest way to measure land, so I would use that.

3 0

3 years ago

Read 2 more answers

Calculate the distance between (4,6) (5,0)

kramer

The distance between two (4, 6) and (5, 0) is 6.08 units.

Solution:

The given points are (4, 6) and (5, 0).

$x_1=4, y_1=6, x_2=5, y_2=0$

To find the distance between two points:

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$d=\sqrt{(5-4)^2+(0-6)^2}$

$d=\sqrt{1^2+(-6)^2}$

The value of 1² = 1

The value of (–6)² = 36

$d=\sqrt{1+36}$

$d=\sqrt{37}$

d = 6.08

The distance between two (4, 6) and (5, 0) is 6.08 units.

5 0

3 years ago

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

b)

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

c)

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

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

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

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

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