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MArishka [77]
3 years ago
6

A simple random sample of 25 filtered 100 mm cigarettes is​obtained, and the tar content of each cigarette is measured. The samp

le has a mean of 19.4 mg and a standard deviation of 3.23 mg. Use a 0.05 significance level to test the claim that the mean tar content of filtered 100 mm cigarettes is less than 21.1 ​mg, which is the mean for unfiltered king size cigarettes.Please select all TRUE statements from those givem below.1. For this hypothesis test the P-value = .0072. There is sufficient evidence to support the claim that the mean tar content of filtered 100 mm cigarettes is less than21.1 mg.3. For this hypothesis test; H0; 1.1 H1: 21.14. For this hypothesis test the test statistic is t = -2.6325. service.php?service=cache&name=55e1f2083
Mathematics
1 answer:
qwelly [4]3 years ago
5 0

Answer:

h0:u ≥ 21.1

h1: u < 21.1

t statistics = -2.632

pvalu = 0.007

there is sufficient evidence to support claim

Step-by-step explanation:

null hypothesis

h0 = u≥21.1

alternative = u < 21.1

to get test statistics

barx - u / s/√n

= 19.4 - 21.1 / 3.23/√25

= -2.632

we calculate for the p value given this test statistics and the degree of freedom = 25-1 = 24

lest tailed p value = 0.0073 this is approximately 0.007

p value is less than the level of significance so we reject h0

we conclude that we have enough evidence to support h1, mean tar conent is less than 21.1

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A rectangle has an area of 24 sq. inches and a perimeter of 50 in. what are the dimensions of the rectangle​
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Answer:

Length of rectangle = 24 inches

Width of rectangle = 1 inches

Step-by-step explanation:

Area of rectangle = 24 square inches

Perimeter of rectangle = 50 inches

Formulas:

Area of rectangle = l\times w

Perimeter of rectangle = 2(l+w)

where l represents length of rectangle and w represents width of rectangle.

So, we can get two equations for l and w

A) l\times w=24    [Area]

B) 2(l+w)=50     [ Perimeter ]

Simplifying equation B.

Dividing both sides by 2.

\frac{2(l+w)}{2}=\frac{50}{2}  

l+w=25  

Solving for l in terms of width.

Subtracting w both sides.

l+w-w=25-w  

∴ l=25-w

Substituting value of l in terms of w in equation A.

(25-w)w=24

Using distribution.

25w-w^2=24

Adding w^2 both sides.

25w-w^2+w^2=24+w^2

25w=24+w^2

subtracting 25 w both sides

25w-25w=24+w^2-25w

0=w^2-25w+24

We have a quadratic equation. w^2-25w+24=0

We solve for w by factor method.

Splitting middle term into two terms such that the sum of the two =-25w and product of two = 24w^2

w^2-24w-w+24=0

Factoring in pairs.

w(w-24)-1(w-24)=0

Factoring as a whole

(w-24)(w-1)=0

So, we have

w-24=0 and w-1=0

By adding 24 to one equation and adding 1 to other we get

w-24+24=0+24 and w-1+1=0+1

w=24 and w=1

So length can be found out by plugging value of w in the length equation.

l=25-24=1 and l=25-1=24

Since length is always the longer side of rectangle so,  the dimensions of rectangle can be given by.

length = 24 inches  (answer)

width = 1 inches (answer)

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