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fgiga [73]
3 years ago
5

A)231*12 531*23 505*11 402*44 b) 516*25 893*72 860*91 c) 413*25 120*518 119*98 7*00*125 d) 7800*8600 980*37000 1860*890 5800*470

0. * = razy
HELP
Mathematics
1 answer:
Allushta [10]3 years ago
5 0

Answer:

mega mind<3<3<3<<3<3?3<3<

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A bucket holds b liters of water. But a jar holds 8 fewer liters than the bucket. How many times can the bucket hold than the ja
kykrilka [37]

Answer:

b / (b - 8)

Step-by-step explanation:

Given that :

Litres of water held by bucket = b

Jar holds 8 litres fewer than bucket; hence, litres held by jar = b - 8

The number of times bucket can hold than jar is :

Litres held by bucket / litres held by jar

b / (b - 8)

8 0
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If Clay can run one lap in a ¼ of an hour. How many hours will it take him to run 5 ⅓ laps around the track.​
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Answer: 21

Step-by-step explanation

5 1/4 divided by 1/4

u simplify 5 1/4 =21/4

1/4=0.25

21/4 divided by 1/4 is 21

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3 years ago
Add the two expressions. −4x+3 and 2x−5
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List 3 words or phrases used in real life that indicate negative numbers.
andrew-mc [135]
A number that has a negative sign in front of it, opposite from positive numbers, and below zero are all phrases that can describe negative numbers. 
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4 years ago
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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
4 years ago
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