All categories

3 years ago

Bonjour j’ai un exercice de maths et je ne comprends rien

1 answer:

4 0

Donc pour trouver la réduction de 30% nous allons trouver 30% de 49 alors nous allons faire 30 × 100 × 49 = 14,7 14,7 est la réduction maintenant nous devons faire 49 € -14,7 € soit 34,3 €, le prix est maintenant de 34,3 €, désolé d'utiliser Google Translate pour vous aider, mais j'espère que c'est correct :)

You might be interested in

Multiply: (49)(58)

IrinaVladis [17]

(49)(58) = 2,842 49 x58 ------ 2,842

4 0

3 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

3 years ago

James buys 2 4/5 kilograms of red potatoes. He buys yellow potatoes that weigh 3/4 as much as the red potatoes.

Illusion [34]

Answer:

Step-by-step explanation:

2 4/5 kg” would equal as a whole number 14/5 kgs, or 2.8, divided by 3/4’s would be 2.1 kgs.

As a mixed number there would be 2 1/10’s.

2 1/10.

8 0

2 years ago

According to a survey, the average American person watches TV for 3 hours per week. To test if the amount of TV in New York City

Neporo4naja [7]

Answer:

Test statistic (t-value) of this one-mean hypothesis test is -2.422.

Step-by-step explanation:

We are given that according to a survey, the average American person watches TV for 3 hours per week. She surveys 19 New Yorkers randomly and asks them about their amount of TV each week, on average. From the data, the sample mean time is 2.5 hours per week, and the sample standard deviation (s) is 0.9 hours.

We have to test if the amount of TV in New York City is less than the national average.

Let Null Hypothesis, $H_0$ : $\mu$ $\leq$ 3 {means that the amount of TV in New York City is less than the national average}