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Schach [20]
3 years ago
6

3x+4y=16 find the value of x and y

Mathematics
2 answers:
GREYUIT [131]3 years ago
7 0
Y= 4- 3x/4
X= 16/3 - 4y/3
zlopas [31]3 years ago
6 0
Y is equal to 4-3x/4 and X=16/3 -4y/3

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A bag contains 1 yellow ball and 9 red balls . A ball is chosen at random from the bag . what is the best answer for the probabi
Strike441 [17]
Your answer is unlikely. That would be a 1 out of 10 chance.
6 0
4 years ago
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!!!

4 0
3 years ago
Help me pls I need help with this test
ValentinkaMS [17]

Answer:

question 7 is A

Step-by-step explanation:

7 0
3 years ago
Oscar has received a $4,000 gift and is looking for a bank to start a savings account. Which option
hoa [83]

Answer:

The best option for him would be a real interest rate of 5%.

Step-by-step explanation:

The nominal interest rate is the one that represents the percentage of increase of the money that is in a certain investment, without discounting the depreciation due to inflation or the payment of taxes.

On the other hand, the real interest rate is the one that represents the real increase in the money invested, after discounting inflation and any taxes to be paid.

Therefore, the best option for Oscar would be to invest his $ 4,000 in a savings account with a real interest rate of 5% per year.

4 0
3 years ago
Intravenous fluid bags are filled by an automated filling machine. Assume that the fill volumes of the bags are independent, nor
kenny6666 [7]

Answer:

a) 0.0167

b) 0

c) 5.948

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.16 ounces

Standard Deviation, σ = 0.08 ounces

We are given that the distribution of fill volumes of bags is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) Standard deviation of 23 bags

\displaystyle\frac{S.D}{\sqrt{23}} = \frac{0.08}{\sqrt{23}} = 0.0167

b) P( fill volume of 23 bags is below 5.95 ounces)

P(x < 5.95)

P( x < 5.96) = P( z < \displaystyle\frac{5.95 - 6.16}{0.0167}) = P(z < -12.57)

= 1 - P(z < 12.57)

Calculation the value from standard normal z table, we have,  

P(x < 5.95) = 1 - 1 = 0

c) P( fill volume of 23 bags is below 6 ounces)  = 0.001

P(x < 6)  = 0.001

P( x < 6) = P( z < \displaystyle\frac{6 - \mu}{0.0167})

Calculation the value from standard normal z table, we have,  

P( z \leq -3.09) = 0.001

\displaystyle\frac{6 - \mu}{0.0167} = -3.09\\\\\mu = 6 + (0.0167\times -3.09) = 5.948

If the mean will be 5.948 then the probability that the average of 23 bags is below 6.1 ounces is 0.001.

7 0
3 years ago
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