2.07)Solve for H: A = L ⋅ W ⋅ H

Angle between two lines What is the size of the angle between the two lines 3x-2y+1=0 and x-3y=0 (answer in degrees and minutes)

adell [148]

Answer:

hiiiiiiiiiiiiiiiiiiiiiiiiii

6 0

2 years ago

Evaluate f(x) = 2x +3 for f(4) a-11 b-8 c-4 d-3

Scrat [10]

Answer:

the answer is a-11

Step-by-step explanation:

f(x)=2x+3

f(4)=2(4)+3

f(4)=8+3

f(4)=11

5 0

2 years ago

Luke is decorating trees for an outdoor birthday party. There are 3 trees and Luke needs 4.25 yards of ribbon for each tree.

suter [353]

Answer:

17.85

Step-by-step explanation:

it would be $5.95 per tree times that by 5 you get $17.85

3 0

3 years ago

Kelly and 23 friends go roller skating they've had a total of $186 about how much does the cost for one person to the skate

pashok25 [27]

about 8 dollars per person

5 0

3 years ago

Read 2 more answers

A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei

Arte-miy333 [17]

Answer:

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

The null hypothesis is accepted .

Assume the population variances are approximately the same

Step-by-step explanation:

Explanation:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

The first sample size 'n₁'= 20

mean of the first sample 'x₁⁻'= 17.53 pounds

standard deviation of first sample S₁ = 3.2 pounds

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

The second sample size n₂ = 24

mean of the second sample "x₂⁻"= 14.89 pounds

standard deviation of second sample S₂ = 2.7 pounds

Null hypothesis:-H₀: The Population Variance are approximately same

Alternatively hypothesis: H₁:The Population Variance are approximately same

Level of significance ∝ =0.05

Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42

Test statistic :-

$t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$

where $S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}$

$S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}$

substitute values and we get S² = 40.988

$t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}$

t = 1.3622

Calculated value t = 1.3622

Tabulated value 't' = 2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

Conclusion:-

The null hypothesis is accepted

Assume the population variances are approximately the same.

6 0

2 years ago