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2 years ago

Solve 2tan(x)cos^2(x)=1 algebraically

Mathematics

1 answer:

deff fn [24]2 years ago

7 0

X= pi/4 +k(pi)
(This may not be correct sorry)

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John buys 100 shares of stock at $100 per share. The price goes up by 10% and he sells 50 shares. Then, prices drop by 10% and h

Juliette [100K]

The profit that John got from the last 50 was $5000.

5 0

3 years ago

In 2010, Josh purchased a house for $110,000. In 2014, the house was valued at $105,000. What is the rate of change for the valu

aniked [119]

22 percent
110000 divided by 22= 5000
110000 - 5000= 105000

6 0

3 years ago

Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%

Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

n = 10

Compute the sample mean and sample standard deviation:

$\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7$

$s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702$

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a t confidence interval.

The critical value of t for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

$t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262$

*Use a t-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}$

$=176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)$

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

7 0

3 years ago

EASY 6TH GRADE QUESTION MATH

Nataliya [291]

Answer: quarts

Step-by-step explanation: 4 cups = 1 quart

6 0

3 years ago

The life in hours of a 75-watt light bulb is known to be normally distributed with standard deviation σ = 25 hours. A random sam

irina [24]

Answer: a) (1008.34,1019.658) b) (1009.24,1018.76)

Step-by-step explanation:

Since we have given that

n = 75

mean = 1014 hours

Standard deviation = 25 hours

At 95% two sided , z = 1.96

So, confidence interval would be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=1014\pm 1.96\dfrac{25}{\sqrt{75}}\\\\=1014\pm 5.658\\\\=(1014-5.658,1014+5.658)\\\\=(1008.34,1019.658)$

(b) Construct a 95% lower confidence bound on the mean life.

z = 1.65

So, confidence interval would be

$\bar{x}\pm z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1014\pm 1.65\times \dfrac{25}{\sqrt{75}}\\\\=1014\pm 4.76\\\\=(1014-4.76,1014+4.76)\\\\=(1009.24,1018.76)$

Hence, a) (1008.34,1019.658) b) (1009.24,1018.76)

4 0

4 years ago