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Semenov [28]
3 years ago
10

Find sin0, cot0, and csc0, where 0 is the angle shown in the figure. Give exact values, not decimal approximations. BRANLIEST!​

Mathematics
1 answer:
Hoochie [10]3 years ago
7 0

Answer:

Sin0 = 25/43

cot0 = 7/5

csc0 = 43/35

Step-by-step explanation:

X^2 = 5^2 + 7^2

X^2 = 25 +49

X^2 = 74

X = sqrt (74) = 8.6

Opp = 5

Adj = 7

Hypo = 8.6

sin0 = opp/Hypo = 5/8.6

Sin0 = 50/86 = 25/43

cot0 = 1/tan0 = adj/opp

cot0 = 7/5

csc0 = 1/cos0 = hypo/adj

csc0 = 8.6/7 = 86/70 = 43/35

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A marketing consultant observed 50 consecutive shoppers at a supermarket. One variable of interest was how much each shopper spe
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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

The mean is the sum of all of the data divided by the number of observations for this case, that is, n = 50.

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.

3 0
3 years ago
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