A conductor is mapping a trip and records the distance the train travels over certain time intervals.

Marvin has a coupon the discount is the rental of a full-size car by $25 they decide to buy insurance for each day if the cost i

ipn [44]

Answer: Marvin needs approximately 19 days to rent the car.

Step-by-step explanation:

Since we have given that

Amount Marvin get on a discount coupon = $25

She decided to buy insurance for each day.

Let the number of days be x

cost of insurance = $465

According to question, it becomes,

$25x=465\\\\x=\frac{465}{25}\\\\x=18.6\\\\x=19\ days$

Hence, Marvin needs approximately 19 days to rent the car.

7 0

1 year ago

Read 2 more answers

A restaurant serves custom-made omelets, where guests select meat, cheese, and vegetables to be added to their omelet. There are

kondor19780726 [428]

Answer: The number of different combinations of 2 vegetables are possible = 15 .

Step-by-step explanation:

In Mathematics , the number of combinations of selecting r values out of n values = $^nC_r=\dfrac{n!}{r!(n-r)!}$

Given : Number of available vegetables = 6

Then, the number of different combinations of 2 vegetables are possible will be :

$^6C_2=\dfrac{6!}{2!(6-2)!}=\dfrac{6\times5\times4!}{2\times4!}=15$

Hence , the number of different combinations of 2 vegetables are possible = 15 .

5 0

2 years ago

mrs bean bought a business at the start of 2006 the business was then valued £30000 Since Mrs Beans takeover, the business gas c

Anna [14]

Answer:

38000(Rounded to nearest 100)

Step-by-step explanation:

2018-2006=12 30000X1.02^12 =38047.2538

4 0

2 years ago

Lori bought a shirt and a hat at a half off sale. If she spent a total of $21 on the two items, what was the original price of t

Bad White [126]

Do 21 divided by 2 and you’ll get 10.5 so the original price was $10 and 50 cents

6 0

1 year ago

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of

Thepotemich [5.8K]

<h2><u>Answer with explanation</u>:</h2>

As per given , we have

sample size : n= 65

degree of freedom : df=n-1=64

sample mean : $\overline{x}=19.5$

sample standard deviation : s= 5.2

Since , the population standard deviation is not given , so we apply t-test.

Significance level for 90% confidence : $\alpha=1-0.90=0.10$

t-critical value for significance level 0.10 and df = 64 would be :

$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.6690$

Formula for Confidence interval :

$\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}$

Then , 90% confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.669)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.076$

$(19.5-1.076,\ 19.5+1.076)=(18.424,\ 20.576)$

∴ 90% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.424, 20.576)

Significance level for 95% confidence : $\alpha=1-0.95=0.05$

t-critical value for significance level 0.05 and df = 64 would be :

$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.9977$

Then , 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.9977)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.288$

$(19.5-1.288,\ 19.5+1.288)=(18.212,\ 20.788)$

∴ 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.212, 20.788)

4 0

2 years ago