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vfiekz [6]
2 years ago
12

Marina has $145 to spend. She spends $68. 50 on jeans. T-shirts cost $12. 75. Write and solve an inequality to find the greatest

number of t-shirts she can buy
Mathematics
1 answer:
Kruka [31]2 years ago
6 0

Answer:

6

Step-by-step explanation:

145-68.50 = 76.50

76.20/12.75 = no of t-shirts she can buy

ans= 6

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Draw a box plot for the following data. {35, 50, 50, 48, 48, 32, 38, 41, 48, 34, 29, 23, 41, 34, 43}
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I can't really draw one, but the points would be, 23,34,41,48,50

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Factor the expression completely. 1/4x + 19/4
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You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confide
FrozenT [24]

Question:

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of ​$114.00. Assume the population standard deviation is ​$15.30. Construct a​ 90% confidence interval for the population mean.

Answer:

At the 90% confidence level, confidence interval = 110.2484 < μ < 117.7516

At the 95% confidence level, confidence interval = 109.53 < μ < 118.48

The 95% confidence interval is wider

Step-by-step explanation:

Here, we have

Sample size, n = 45

Sample mean, \bar x = $114.00

Population standard deviation, σ = $15.30

The formula for Confidence Interval, CI is given by the following relation;

CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}

Where, z is found for the 90% confidence level as ±1.645

Plugging in the values, we have;

CI=114\pm 1.645 \times \frac{15.3}{\sqrt{45}}

or CI: 110.2484 < μ < 117.7516

At 95% confidence level, we have our z value given as z = ±1.96

From which we have CI=114\pm 1.96 \times \frac{15.3}{\sqrt{45}}

Hence CI: 109.53 < μ < 118.48

To find the wider interval, we subtract their minimum from the maximum as follows;

90% Confidence level: 117.7516 - 110.2484 = 7.5

95% Confidence level: 118.47503 - 109.5297 = 8.94

Therefore, the 95% confidence interval is wider.

8 0
3 years ago
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Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering p
tankabanditka [31]

Answer:

1.778 times more or 16/9 times more

Step-by-step explanation:

Given:

- Mirror 1: D_1 = 8''

- Mirror 2: D_2 = 6"

Find:

Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering power?

Solution:

- The light gathering power of a mirror (LGP) is proportional to the Area of the objects:

                                           LGP ∝ A

- Whereas, Area is proportional to the squared of the diameter i.e an area of a circle:

                                           A ∝ D^2

- Hence,                              LGP ∝ D^2

- Now compare the two diameters given:

                                           LGP_1 ∝ (D_1)^2

                                           LGP ∝ (D_2)^2

- Take a ratio of both:

                           LGP_1/LGP_2 ∝ (D_1)^2 / (D_2)^2

- Plug in the values:

                               LGP_1/LGP_2 ∝ (8)^2 / (6)^2

- Compute:             LGP_1/LGP_2 ∝ 16/9 ≅ 1.778 times more

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Answer:

Step-by-step explanation:

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