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miv72 [106K]
3 years ago
10

Haley is comparing the cost of a fresh lobster dinner at two different restaurants. The first restaurant charges $29 for the mea

l, plus $6 per kilogram for the lobster she picks. At the second restaurant, Haley would pay $3 per kilogram for the lobster, in addition to $41 for the meal. Haley realizes that, in theory, dinner at both restaurants could cost the same amount if the lobster had a certain weight. How much would Haley pay for her dinner? What is the weight?
Mathematics
1 answer:
Svetllana [295]3 years ago
8 0

First restaurant price: 29 + 6l = x

x = final cost

l = kilo of lobster

Second restaurant price: 41 + 3l = x

We want them to be the same price, so: 29 + 6l = 41 + 3l

Subtract 3l from each side

29 + 3l = 41

Subtract 29 from both sides

3l = 12

Divide both sides by 3

l = 4

So, if she buys 4 kilograms of lobster from either place, it will cost the same.

It would cost $53.

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NeX [460]

Answer:

39.25

Step-by-step explanation:

the formula is a = (3.14 × r^{2}) / 2

the equation is a = (3.14 ×5^{2}) / 2

5 is the radius

1. 5^{2} = 25

2. 3.14 × 25 = 78.5

3. 78.5 / 2 = 39.25

I hope this helped! :)

3 0
3 years ago
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Solve for y. <br><br> 15 points to anyone who answers. Seriously need help with this!!!
gtnhenbr [62]
The answer to it is x=12
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3 years ago
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f a sampling distribution is created using samples of the amounts of weight lost by 51 people on this diet, what would be the st
galina1969 [7]

This question is incomplete, the complete question is;

A study on the latest fad diet claimed that the amounts of weight lost by all people on this diet had a mean of 24.6 pounds and a standard deviation of 8.0 pounds.

Step 2 of 2 : If a sampling distribution is created using samples of the amounts of weight lost by 51 people on this diet, what would be the standard deviation of the sampling distribution of sample means? Round to two decimal places, if necessary.

Answer:

the standard deviation of the sampling distribution of sample means is 1.12

Step-by-step explanation:

Given the data in the question,

population mean; μ = 24.6  pounds

Population standard deviation; σ = 8.0 pounds

sample size; n = 51

Now determine the standard deviation of the sampling distribution of sample means.

standard deviation of the sampling distribution of sample means is simply

⇒ population standard deviation / √sample size

= 8.0 / √51

= 8.0 / 7.141428

= 1.120224 ≈ 1.12

Therefore, the standard deviation of the sampling distribution of sample means is 1.12

5 0
2 years ago
g A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is cha
leva [86]

Answer:

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}  

z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502    

p_v =2*P(Z>2.502)= 0.0123    

Comparing the p value with the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.    

Step-by-step explanation:

Data given and notation    

X_{1}=410 represent the number of people indicating that their financial security was more than fair for the recent year

X_{2}=245 represent the number of people indicating that their financial security was more than fair for the year before

n_{1}=1000 sample 1 selected  

n_{2}=700 sample 2 selected  

p_{1}=\frac{410}{1000}=0.410 represent the proportion estimated of indicating that their financial security was more than fair this year

p_{2}=\frac{245}{700}=0.35 represent the proportion estimated of indicating that their financial security was more than fair the year before

\hat p represent the pooled estimate of p

z would represent the statistic (variable of interest)    

p_v represent the value for the test (variable of interest)  

\alpha significance level given  

Concepts and formulas to use    

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

We need to apply a z test to compare proportions, and the statistic is given by:    

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{410+245}{1000+700}=0.385  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.    

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:    

z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502    

Statistical decision  

Since is a two sided test the p value would be:    

p_v =2*P(Z>2.502)= 0.0123    

Comparing the p value with the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.    

7 0
3 years ago
What is 528 1/5 - 89 3/5
sergiy2304 [10]
The answer is 438.6...
(528+1/5)-(89+3/5)=438.6
8 0
3 years ago
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