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Ivan
3 years ago
6

Uhhh I am struggling with this question algebra 1

Mathematics
1 answer:
Olin [163]3 years ago
4 0

Answer:

-2x^2 +16x -8

Step-by-step explanation:

Oh s-

Like terms i guess Same variable you can substract. Remember you only substract from 1 number not from all.

-2x^2 stays the same as there is no like term

9x - (-7x)

-2 - (6)

-2x^2 + 9x+ 7x -2-6

-2x^2 +16x -8

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Hello This is pretty Easy But i forgot the basics
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Answer:

-3, -2 1/2, -2 2/5, -2 3/10, -2

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3 years ago
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A random sample of 20 recent weddings in a country yielded a mean wedding cost of $ 26,388.67. Assume that recent wedding costs
Makovka662 [10]

Answer:

a) 95% confidence interval for the mean​cost, μ​, of all recent weddings in this country = (22,550.95, 30,226.40)

.The​ 95% confidence interval is from $22,550.95 to $30,226.40.

b) For the interpretation of the result, option D is correct.

We can be​ 95% confident that the mean​ cost, μ​, of all recent weddings in this country is somewhere within the confidence interval.

c) Option B is correct.

The population mean may or may not lie in this​ interval, but we can be​ 95% confident that it does.

Step-by-step explanation:

Sample size = 20

Sample Mean = $26,388.67

Sample Standard deviation = $8200

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 26,388.67

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 20 - 1 = 19.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 19) = 2.086 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 8200

n = sample size = 20

σₓ = (8200/√20) = 1833.6

99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 26,388.67 ± (2.093 × 1833.6)

CI = 26,388.67 ± 3,837.7248

99% CI = (22,550.9452, 30,226.3948)

99% Confidence interval = (22,550.95, 30,226.40)

a) 95% confidence interval for the mean​cost, μ​, of all recent weddings in this country = (22,550.95, 30,226.40)

.The​ 95% confidence interval is from $22,550.95 to $30,226.40.

b) The interpretation of the confidence interval obtained, just as explained above is that we can be​ 95% confident that the mean​ cost, μ​,of all recent weddings in this country is somewhere within the confidence interval

c) A further explanation would be that the population mean may or may not lie in this​ interval, but we can be​ 95% confident that it does.

Hope this Helps!!!

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2 years ago
Solve for x (5x+15)(3x-3)
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You will need to find how much he got on July so you will need to multiply 1.10 and 31.

1.10 time 31 is 34.1

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Answer:6

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