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Mamont248 [21]
2 years ago
7

Today, 6 friends went out for lunch. Their total bill was $27.60, including tax and gratuity. They decided to split the bill equ

ally and each paid with a $10 bill. How much money will each person get back?
Mathematics
2 answers:
Paladinen [302]2 years ago
7 0

Step-by-step explanation:

Right, okay, so this is a bit of a weird one. Let's go step by step.

"Today, 6 friends went out for lunch." = The number 6 is gonna be in our equation somewhere.

"Their total bill was $27.60" = so is the number 27.6

"They decided to split the bill equally" = 27.6 "split equally" (aka divided by) 6 friends is 4.6

"each paid with a $10 bill" = This is where it gets a lil weird. So, if they divided the bill among them, none of their amounts are gonna be anywhere near ten dollars. This is one of those questions that wouldn't really happen in real life. People would probably use fives instead, but whatever. So a number we'll be using for something is 10.

"How much money will each person get back?" = So we have to find the amount they all paid (10) minus the amount each one had to pay (4.6).

To put it all into a full equation...

10 - ( 27.6 / 6 )

Divide.

10 - 4.6

Subtract.

5.4

Put back into money form.

$5.40

Answer:

Each person will get back $5.40.

Liula [17]2 years ago
4 0

Answer:

if each paid a $10 bill, in total they paid $60 (cuz 10 x 6)

60 - 27.60 = 32.4

32.4/6 = 5.4 (divide the change between 6 people)

Each person will get $5.40 back.

<em>hope it helps :)</em>

<em />

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(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (<em>n</em> ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

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Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18

The sample size is, <em>n</em> = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

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\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191

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<em>H</em>₀: The usual load is 13 credits, i.e. <em>μ</em> = 13.

Assume that the significance level of the test is, <em>α</em> = 0.05.

Construct a (1 - <em>α</em>) % confidence interval for population mean to check the claim.

The (1 - <em>α</em>) % confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\times SE

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CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)

As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

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Compute the probability that a a student at this college takes 16 or more credits as follows:

P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

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