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Step2247 [10]
3 years ago
11

PLEASE HELP WILL GIVE BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLI

EST BRAINLIEST BRAINLIEST (MERRY CHRISTMAS!!!!)

Mathematics
2 answers:
Alex17521 [72]3 years ago
5 0

Answer:

4+2=6

15*6=90

9+9=18

18*3=54

90+54=144

Step-by-step explanation:

you have to add both the W and H together and then multiply them by the number of windows (4+2=6 , 16*6=90)

and the same with the garage doors (9+9=18, 18*3=54) then you take you two totals from multiplying and add them together (90+54=144)

Rasek [7]3 years ago
4 0

Answer:

8*8*8*8*8*8*8*8*8*8*8*8*8*8*8*8*8*8*8*8*8*8*8*8*=216

Step-by-step explanation:

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A bag contains green yellow and black marbles out of the 100 marbles in the bag 35 marbles are green and 15 or yellow what perce
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Step-by-step explanation:

well, there are 100 marbles

there are 35 and 15 already known which equals to 50

therefore 50% of the bag is black marbles

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The answer of this question is b
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77

Step-by-step explanation:

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Up or down and maximum or minimum
just olya [345]
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3 years ago
Two machines are used for filling plastic bottles to a net volume of 16.0 ounces. A member of the quality engineering staff susp
aleksandrvk [35]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.

The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.

You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:

Sample 1 - Machine 1

X₁: Net volume of a plastic bottle filled by the machine 1.

n₁= 10

X[bar]₁= 16.02

S₁= 0.03

Sample 2 - Machine 2

X₂: Net volume of a plastic bottle filled by the machine 2

n₂= 10

X[bar]₂= 16.01

S₂= 0.03

With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:

X₁~N(μ₁;δ₁²)

X₂~N(μ₂;δ₂²)

Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)

H₀: δ₁²/δ₂²=1

H₁: δ₁²/δ₂²≠1

α: 0.05

F= \frac{S^2_1}{S^2_2} * \frac{Sigma^2_1}{Sigma^2_2} ~~F_{n_1-1;n_2-1}

Using a statistic software I've calculated the test

F_{H_0}= 1.41

p-value 0.6168

The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.

To study the difference between the population means you can use

t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~~t_{n_1+n_2-2}

The hypotheses of interest are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

Sa= \sqrt{\frac{(n_1-1)S^2_1+(N_2-1)S^2_2}{n_1+n_2-2}} = \sqrt{\frac{9*9.2*10^{-4}+9*6.5*10^{-4}}{10+10-2} } = 0.028= 0.03

t_{H_0}= \frac{(16.02-16.01)-0}{0.03*\sqrt{\frac{1}{10} +\frac{1}{10} } } = 0.149= 0.15

The p-value for this test is: 0.882433

The p-value is greater than the significance level, the decision is to not reject the null hypothesis.

Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.

I hope this helps!

8 0
3 years ago
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