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3 years ago

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Use technology to help you test the claim about the population mean, mu, at the given level of significance, alpha, using the gi

ven sample statistics. Assume the population is normally distributed.
Claim: μ>1220;α=0.08; σ=211.67.
Sample statistics: x=1235.91,n=300
Identify the null and alternative hypotheses and calculate the standardized test statistic.

1 answer:

Sergio [31]3 years ago

3 0

Answer:

H0 : μ = 1220

H1 : μ > 1220

Test statistic = 1.30

Step-by-step explanation:

Sample mean, x = 1235.91

Standard deviation, σ = 211.67

Sample size, n = 300

The hypothesis :

Null ; H0 : μ = 1220

Alternative ; H1 : μ > 1220

Tbe test statistic :

(x - μ) ÷ (σ/√(n))

(1235.91 - 1220) ÷ (211.67/√(300))

15.91 / 12.220773

= 1.3018

= 1.30

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Answer:

I) If method I is used, population of generalization will include all those people who may have varying exercising habits or routines. They may or may not have a regular excersing habit. In his case sample is taken from a more diverse population

II) Population of generalization will include people who will have similar excersing routines and habits if method II is used since sample is choosen from a specific population

Step-by-step explanation:

past excercising habits may affect the change in intensity to a targeted excersise in different manner. So in method I a greater diversity is included and result of excersing with or without a trainer will account for greater number of variables than method II.

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The equation 2x2 − 12x + 1 = 0 is being rewritten in vertex form. Fill in the missing step. Given 2x2 − 12x + 1 = 0 Step 1 2(x2

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Answer:

Part 1) $2(x-3)^{2}-17=0$ (the missing steps in the explanation)

Part 3) (8, 4); The vertex represents the maximum profit

Part 4) x = 3.58, 0.42

Part 5) x = 6, x = 44; The zeros represent the number of monthly memberships where no profit is made

Part 6) 2(x − 7)2 + 118; x = $7

Part 7) The maximum height of the puck is 4 feet. −(x − 4)^2 + 6

Part 8) (x + 3)^2 − 4

Part 9) 2(x − 1)^2 = 4

Part 10) 8(x − 4)^2 + 592

Step-by-step explanation:

Part 1) we have

$2x^{2} -12x+1=0$

Convert to vertex form

step 1

Factor the leading coefficient and complete the square

$2(x^{2} -6x)+1=0$

$2(x^{2} -6x+9)+1-18=0$

step 2

$2(x^{2} -6x+9)+1-18=0$

$2(x^{2} -6x+9)-17=0$

step 3

Rewrite as perfect squares

$2(x-3)^{2}-17=0$

Part 3) we have

$f(x)=-x^{2}+16x-60$

we know that

This is the equation of a vertical parabola open downward

The vertex is a maximum

Convert to vertex form

$f(x)+60=-x^{2}+16x$

Factor the leading coefficient

$f(x)+60=-(x^{2}-16x)$

Complete the squares

$f(x)+60-64=-(x^{2}-16x+64)$

$f(x)-4=-(x^{2}-16x+64)$

Rewrite as perfect squares

$f(x)-4=-(x-8)^{2}$

$f(x)=-(x-8)^{2}+4$

The vertex is the point (8,4)

The vertex represent the maximum profit

Part 4) Solve for x

we have

$-2(x-2)^{2}+5=0$

$-2(x-2)^{2}=-5$

$(x-2)^{2}=2.5$

square root both sides

$(x-2)=(+/-)1.58$

$x=2(+/-)1.58$

$x=2(+)1.58=3.58$

$x=2(-)1.58=0.42$

Part 5) we have

$f(x)=-x^{2}+50x-264$

we know that

The zeros or x-intercepts are the value of x when the value of the function is equal to zero

so

In this context the zeros represent the number of monthly memberships where no profit is made

To find the zeros equate the function to zero

$-x^{2}+50x-264=0$

$-x^{2}+50x=264$

Factor -1 of the leading coefficient

$-(x^{2}-50x)=264$

Complete the squares

$-(x^{2}-50x+625)=264-625$

$-(x^{2}-50x+625)=-361$

$(x^{2}-50x+625)=361$

Rewrite as perfect squares

$(x-25)^{2}=361$

square root both sides

$(x-25)=(+/-)19$

$x=25(+/-)19$

$x=25(+)19=44$

$x=25(-)19=6$

Part 6) we have

$-2x^{2}+28x+20$

This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

Factor the leading coefficient

$-2(x^{2}-14x)+20$

Complete the square

$-2(x^{2}-14x+49)+20+98$

$-2(x^{2}-14x+49)+118$

Rewrite as perfect square

$-2(x-7)^{2}+118$

The vertex is the point (7,118)

therefore

The video game price that produces the highest weekly profit is x=$7

Part 7) we have

$f(x)=-x^{2}+8x-10$

Convert to vertex form

$f(x)+10=-x^{2}+8x$

Factor -1 the leading coefficient

$f(x)+10=-(x^{2}-8x)$

Complete the square

$f(x)+10-16=-(x^{2}-8x+16)$

$f(x)-6=-(x^{2}-8x+16)$

Rewrite as perfect square

$f(x)-6=-(x-4)^{2}$

$f(x)=-(x-4)^{2}+6$

The vertex is the point (4,6)

therefore

The maximum height of the puck is 4 feet.

Part 8) we have

$x^{2}+6x+5$

Convert to vertex form

Group terms

$(x^{2}+6x)+5$

Complete the square

$(x^{2}+6x+9)+5-9$

$(x^{2}+6x+9)-4$

Rewrite as perfect squares

$(x+3)^{2}-4$

Part 9) we have

$2x^{2}-4x-2=0$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 2 the leading coefficient

$2(x^{2}-2x)-2=0$

Complete the square

$2(x^{2}-2x+1)-2-2=0$

$2(x^{2}-2x+1)-4=0$

Rewrite as perfect squares

$2(x-1)^{2}-4=0$

$2(x-1)^{2}=4$

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Part 10) we have

$8x^{2}-64x+720$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 8 the leading coefficient

$8(x^{2}-8x)+720$

Complete the square

$8(x^{2}-8x+16)+720-128$

$8(x^{2}-8x+16)+592$

Rewrite as perfect squares

$8(x-4)^{2}+592$

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It’s simple

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Answer:

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Step-by-step explanation:

nth term of the sequence =n^2 + 20

an= n^2 + 20

1st term when n= 1

1^2 + 20= 20

2nd term n= 2

2^2 + 20=24

3rd term when n= 3

3^2 + 20= 29

4th term when n= 4

4^2 + 20= 36

5th term when n= 5

5^2 + 20 =45

6th term when n= 6

6^2 + 20=56

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