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2 years ago

6

Joseph’s lunch at a restaurant costs $13.00, without tac. He leaves the waiter a tip of 17% of the cost of the lunch, without ta

x. what is the total cost of the lunch, including the tip, without tax?
A) $2.21
B)$10.78
C)13.17
D)15.21

2 answers:

KonstantinChe [14]2 years ago

3 0

The answer is D

17% of 13 is 2.21

13 + 2.21 = 15.21

Kryger [21]2 years ago

3 0

Answer:

Option D

Step-by-step explanation:

Joseph's lunch costs = $13 without taxes

He leaves tip for the waiter = 17% of cost of the lunch

= 17% of $13

= 13 × $\frac{17}{100}$

= ( 13 × 0.17)

= $2.21

Total cost of lunch + tip = 13 + 2.21

= $15.21

Option D is the answer.

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An urn contains six chips, numbered 1 through 6. two are chosen at random and their numbers are added together. what is the prob

vitfil [10]

<span>2/15 if drawn without replacement. 1/9 if drawn with replacement. Assuming that the chips are drawn without replacement, there are 6 * 5 different possibilities. And that's a low enough number to exhaustively enumerate them. So they are: 1,2 : 1,3 : 1,4 : 1,5 : 1,6 2,1 : 2,3 : 2,4 : 2,5 : 2,6 3,1 : 3,2 : 3.4 : 3,5 : 3,6 4,1 : 4,2 : 4.3 : 4,5 : 4,6 5,1 : 5,2 : 5.3 : 5,4 : 5,6 6,1 : 6,2 : 6.3 : 6,4 : 6,5 Of the above 30 possible draws, there are 4 that add up to 5. So the probability is 4/30 = 2/15 If the draw is done with replacement, then there are 36 possible draws. Once again, small enough to exhaustively list, they are: 1,1 : 1,2 : 1,3 : 1,4 : 1,5 : 1,6 2,1 : 2,2 : 2,3 : 2,4 : 2,5 : 2,6 3,1 : 3,2 : 3,3 : 3.4 : 3,5 : 3,6 4,1 : 4,2 : 4.3 : 4,4 : 4,5 : 4,6 5,1 : 5,2 : 5.3 : 5,4 : 5,5 : 5,6 6,1 : 6,2 : 6.3 : 6,4 : 6,5 : 6,6 And of the above 36 possibilities, exactly 4 add up to 5. So you have 4/36 = 1/9</span>

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2 years ago

The first term of a finite geometric series is 6 and the last term is 4374. The sum of all the term os 6558. find the common rat

Oliga [24]

A geometric series is written as $ar^n$ , where $a$ is the first term of the series and $r$ is the common ratio.

In other words, to compute the next term in the series you have to multiply the previous one by $r$ .

Since we know that the first time is 6 (but we don't know the common ratio), the first terms are

$6, 6r, 6r^2, 6r^3, 6r^4, 6r^5, \ldots$ .

Let's use the other information, since the last term is $4374 > 6$ , we know that $r>1$ , otherwise the terms would be bigger and bigger.

The information about the sum tells us that

$\displaystyle \sum_{i=0}^n 6r^i = 6\sum_{i=0}^n r^i = 6558$

We have a formula to compute the sum of the powers of a certain variable, namely

$\displaystyle \sum_{i=0}^n r^i = \cfrac{r^{n+1}-1}{r-1}$

So, the equation becomes

$6\cfrac{r^{n+1}-1}{r-1} = 6558$

The only integer solution to this expression is $n=6, r=3$ .

If you want to check the result, we have

$6+6*3+6*3^2+6*3^3+6*3^4+6*3^5+6*3^6 = 6558$

and the last term is

$6*3^6 = 4374$

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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

For which discriminant is the graph possible<br> b2-4ac=0<br><br> b2-4ac=-1<br><br> b2-4ac=4

VARVARA [1.3K]

Answer:

The graph is possible for $b^2-4ac=4$

Step-by-step explanation:

we know that

The discriminant of a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$D=b^2-4ac$

If D=0 the quadratic equation has only one real solution

If D>0 the quadratic equation has two real solutions

If D<0 the quadratic equation has no real solution (complex solutions)

In this problem , looking at the graph, the quadratic equation has two real solutions (the solutions are the x-intercepts)

so

$b^2-4ac > 0$

therefore

The graph is possible for $b^2-4ac=4$

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1 year ago

A hospital bill is estimated to be $462.00. It ends up actually costing the patient $525.00. What is the percent error in the bi

Mamont248 [21]

In order to find the percent error, we need to first find the difference between what was expected and what is actually costed. We do this by subtracting:

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So now we know that the expected amount was off by $63. To find the percent error, we need to take this $63, and divide it by the amount that was estimated. Let's do that now:

$\frac{63}{462}=0.1364$

However this is in decimal form. We need to multiply by 100 in order to get it in a percent:

$0.1364*100=13.64%$

Now we know that the percent error of the hospital bill estimate is 13.64%.

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2 years ago