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3 years ago

How to do this helppp ! i need this for my finals

Mathematics

1 answer:

ruslelena [56]3 years ago

7 0

Answer:

9. 168

10.

$3 {x}^{2} - 14x - 24$

Step-by-step explanation:

on the picture

if it's helpful ❤❤❤

THANK YOU.

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The distance between P and T on the coordinate grid is units. (Input whole numbers only.)

nlexa [21]

Answer:ummm do you have a picture of the question?

Step-by-step explanation:

7 0

3 years ago

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Write an inequality to represent the following statement.

harina [27]

X<2.5

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5 0

3 years ago

Is ∆RST ≅ ∆VUT ? Justify your answer below.

BlackZzzverrR [31]

Answer:

Yes.

Step-by-step explanation:

We know <S is congruent to <U. line segment ST is congruent to line segment UT. Also, the line going through T( line segment RV) makes 2 congruent angles. <STU and < UTV are congruent. So triangle RST is congruent to triangle VUT by ASA (angle side angle.)

8 0

3 years ago

Which two ratios form a proportion A) 1/2 and 2/7 B) 2/1 and 7/14 C) 1/2 and 7/14 D) 2/1 and 14/21

Romashka [77]

1/2 * 2/2 = 2/4 which is not 2/7

2/1 * 7/7 = 14/7 which is not 7/14

1/2 * 7/7 = 7/14 which is 7/14 yes

2/1 * 7/7 = 14/7 which is not 14/21 no

Choice C

4 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

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