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rosijanka [135]
4 years ago
13

The length of a rectangle is 2 cm longer than its width. if the perimeter of the rectangle is 40 cm , find its area.

Mathematics
1 answer:
rusak2 [61]4 years ago
4 0

width = w

 length = w +2

perimeter = 2L + 2w

40 = 2(w+2) + 2w

40 = 2w+4 +2w

40 = 4w+4

36=4w

w=36/4 = 9

with = 9

length = 9+2 = 11


 area = l x w = 9*11 = 99 square cm

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If a1 = 8 and an = an-1 + 5 then find the value of a5.
Nimfa-mama [501]

Step-by-step explanation:

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A2=A1+5 plug 8 into A1 here. After getting the value, put into next equation. And repeat until you get A5

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3 years ago
A marketing consultant observed 50 consecutive shoppers at a supermarket. One variable of interest was how much each shopper spe
Diano4ka-milaya [45]

Answer:

a. \\ \mu = USD\;51.34; b. The 95% interval is USD [45.82, 56.86].

Step-by-step explanation:

The mean amount a customer spends at the grocery store

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The sum of all of the data is: 5.24+11.97+13.59+15.52+15.69+19.96+20.54+22.59+27.25+27.29+28.25+30.00+31.63+33.01+37.21+40.36+40.90+42.83+45.66+46.67+47.49+49.32+50.12+51.71+52.2 1+53.57+56.16+56.36+56.88+58.65+59.42+60.20+60.48+62.23+62.50+67.95+70.45+73.08+74.15+74.48+74.78+75.75+76.27+76.47+78.92+84.09+84.91+85.06+88.45+88.91 = 2567.17.

The mean \\ \mu is

\\ \mu = \frac{2567.17}{50}

\\ \mu = \frac{2567.17}{50} = 51.34

Then, <em>the mean amount a customer spends at the grocery store is</em> USD 51.34.

The 95% confidence interval for the mean amount spent

For the 95% confidence interval for the mean, we have the value for the <em>mean = </em>51.34, as we have just found, the value for the<em> standard deviation</em> is already given as USD 20.00. There are 50 observations, so n = 50.

Moreover, for a 95% confidence, the value for the corresponding z-score is z = 1.96 (this is the confidence coefficient). Then, the formula for finding the "limits" values for the confidence interval is as follows:

For the upper limit  

\\ \overline{X} + 1.96*\frac{\sigma}{\sqrt{n}}

\\ 51.34 + 1.96*\frac{20}{\sqrt{50}}

\\ 51.34 + 5.52

\\ 56.86

For the lower limit

\\ \overline{X} - 1.96*\frac{\sigma}{\sqrt{n}}

\\ 51.34 - 1.96*\frac{20}{\sqrt{50}}

\\ 51.34 - 5.52

\\ 45.82

Thus

The 95% confidence is USD [45.82, 56.86].

Mathematically

\\ P(45.82 \leq \mu \leq 56.86) = 0.95

There is a probability of 0.95 that the values are between [45.82, 56.86], both inclusive.

We have to remember here that we are dealing with a sample. This sample has a standard deviation, called standard error of the sample, and is \\ \frac{20}{\sqrt{50}} \approx 2.83.

So, the population mean is, with a 95% confidence, between the values [45.82, 56.86], both inclusive.

We can see the graph below that represents the former.

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Usimov [2.4K]

Answer:

x ≤ 9

Step-by-step explanation:

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