The product of a number and nine is added to five. The result is thirty-

Please help!!!!!!!!!!!!!!!!!!!!!

dalvyx [7]

30,12,6,6,36 there you go

8 0

4 years ago

What is -3 * 5 + 5 equal

anzhelika [568]

Answer:

-10

Step-by-step explanation:

-15 + 5

-10

4 0

4 years ago

Read 2 more answers

Which of the following functions is not linear?

pickupchik [31]

Answer:

third bubble, y = (1/3)^x

Step-by-step explanation:

8 0

3 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 meancost, 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 meancost, 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

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally

rusak2 [61]

Answer:

(a) The probability that the thickness is less than 3.0 mm is 0.119.

(b) The probability that the thickness is more than 7.0 mm is 0.119.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is 0.762.

Step-by-step explanation:

We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.

Let X = thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.

So, X ~ Normal( $\mu=5.0,\sigma^{2} =1.7^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean thickness = 5.0 mm

$\sigma$ = standard deviation = 1.7 mm

(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

P(X < 3.0 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{3.0-5.0}{1.7}$ ) = P(Z < -1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

P(X > 7.0 mm) = P( $\frac{X-\mu}{\sigma}$ > $\frac{7.0-5.0}{1.7}$ ) = P(Z > 1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X $\leq$ 3.0 mm)

P(X < 7.0 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{7.0-5.0}{1.7}$ ) = P(Z < 1.18) = 0.881

P(X $\leq$ 3.0 mm) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{3.0-5.0}{1.7}$ ) = P(Z $\leq$ -1.18) = 1 - P(Z < 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.

4 0

4 years ago