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Art [367]
2 years ago
7

Joe's hot dog stand is only open for lunch. of today's sales 26% came from selling plain hot dogs. Since the hot dog stand made

$45.50 from selling plain hot dogs, it made $___ in total sales for the day.
Mathematics
1 answer:
meriva2 years ago
6 0
97 then add and find your multiplying

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At a particular restaurant, each onion ring has 70 calories and each slider has 200 calories. A combination meal with onion ring
crimeas [40]

Answer:

3x + 6y

x = sliders

y = onion rings

Step-by-step explanation:

2:1 ratio of onion rings to sliders, add them 200+70+70 = 340,

then 1020/340 = 3, then input into an equation remembering the ratio of onion rings to sliders

7 0
2 years ago
In which set do all of the values make the inequality 2x - 1 < 10 true?
igomit [66]

2x - 1 < 10\Leftrightarrow 2x < 11\Leftrightarrow x <  \frac{11}{2}

8 0
2 years ago
Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of
nikitadnepr [17]

Answer:

a) Reject H₀

b) [0.31; 3.35]

Step-by-step explanation:

Hello!

a) The objective of this example is to compare if the population means of the production rate of the assembly lines A, B and C. To do so the data of the production of each line were recorded and an ANOVA was run using it.

The study variable is:

Y: Production rate of an assembly line.

Assuming that the study variable has a normal distribution for each population, the observations are independent and the population variances are equal, you can apply a parametric ANOVA with the hypothesis:

H₀ μ₁= μ₂= μ₃

H₁: At least one of the population means is different from the others

Where:

Population 1: line A

Population 2: line B

Population 3: line C

α: 0.01

This test is always one-tailed to the right. The statistic is the Snedecor's F, constructed as the MSTr divided by the MSEr if the value of the statistic is big, this means that there is a greater variance due to the treatments than to the error, this means that the population means are different. If the value of F is small, it means that the differences between populations are not significant ( may differ due to error and not treatment).

The critical region is:

F_{k-1;n-k; 1-\alpha } = F_{2;15; 0.99} = 6.36

If F ≥ 3.36, the decision is to reject the null hypothesis.

Looking at the given data:

F= \frac{MSTr}{MSEr}= 11.32653

With this value the decision is to reject the null hypothesis.

Using the p-value method:

p-value: 0.001005

α: 0.01

The p-value is less than the significance level, the decision is to reject the null hypothesis.

At a level of 5%, there is significant evidence to say that at least one of the population means of the production ratio of the assembly lines A, B and C is different than the others.

b) In this item, you have to stop paying attention to the production ratio of the assembly line A to compare the population means of the production ratio of lines B and C.

(I'll use the same subscripts to be congruent with part a.)

The parameter to estimate is μ₂ - μ₃

The populations are the same as before, so you can still assume that the study variables have a normal distribution and their population variances are unknown but equal. The statistic to use under these conditions, since the sample sizes are 6 for both assembly lines, is a pooled-t for two independent variables with unknown but equal population variances.

t=  (X[bar]₂ - X[bar]₃) - ( μ₂ - μ₃) ~t_{n_2+n_3-2}

Sa√(1/n₂+1/n₃)

The formula for the interval is:

(X[bar]₂ - X[bar]₃) ± t_{n_2+n_3-2; 1 - \alpha /2}* Sa\sqrt{*\frac{1}{n_2} + \frac{1}{n_3} }

Sa^{2} = \frac{(n_2-1)*S_2^2+ (n_3-1)*S_3^2}{n_2+n_3-2}

Sa^{2} = \frac{(5*0.67)+ (5*0.7)}{6+6-2}

Sa^{2} = 0.685

Sa= 0.827 ≅ 0.83

t_{n_2+n_3-2;1-\alpha /2}= t_{10;0.995} = 3.169

X[bar]₂ = 43.33

X[bar]₃ = 41.5

(43.33-41.5) ± 3.169 * *0.83\sqrt{*\frac{1}{6} + \frac{1}{6} }

1.83 ± 3.169 * 0.479

[0.31; 3.35]

With a confidence level of 99% you'd expect that the difference of the population means of the production rate of the assemly lines B and C.

I hope it helps!

8 0
2 years ago
One division of a large defense contractor manufactures telecommunication equipment for the military. This division reports that
Zielflug [23.3K]

Answer:

At the 95% confidence level, management should not conclude that the percentage of rework for electrical components is lower than the rate of 12% for non-electrical components, since the upper bound of the confidene interval, which is 0.1339, is higher than 0.12.

Step-by-step explanation:

To reach a conclusion, we have to observe the confidence interval.

Should management conclude that the percentage of rework for electrical components is lower than the rate of 12% for non-electrical components?

Is the upper bound of the confidence interval lower than 12% = 0.12?

If yes, it should conclude that this percentage is lower.

Otherwise, it cannot conclude.

Confidence interval:

0.0758 to 0.1339

0.1339 is higher than 0.12.

So.

At the 95% confidence level, management should not conclude that the percentage of rework for electrical components is lower than the rate of 12% for non-electrical components, since the upper bound of the confidene interval, which is 0.1339, is higher than 0.12.

6 0
2 years ago
Read 2 more answers
A portfolio manager summarizes the input from the macro and micro forecasters in the following table
jasenka [17]
The awnser to this question is Macro Forecasts
5 0
2 years ago
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