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4 years ago

HELP probability Q! Check my answer!

Mathematics

1 answer:

Scrat [10]4 years ago

5 0

A. 3 days a week: 12/80=0.15
B. MORE than 4 days a week: 28/80= 0.4625
C. at LEAST once a week: 0.8125

Correct.

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HELP

Amiraneli [1.4K]

Answer:

volume of a cone = 1/3*(πr²h)

Step-by-step explanation:

6 0

2 years ago

The concentration of active ingredient in a liquid laundry detergent is thought to be affected by the type of catalyst used in t

alisha [4.7K]

Answer:

a. The 95% C.I. is approximately -5.83 < μ₁ - μ₂ < -0.57

b. Yes

c. The power of the test is 0.03836

d. The sample size can be considered adequate

Step-by-step explanation:

a. The given data in the study

Catalyst 1: 57.9, 66.2, 65.4, 65.4, 65.2, 62.6, 67.6, 63.7, 67.2, 71.0

Catalyst 2: 66.4 , 71.7, 70.3, 69.3, 64.8, 69.6, 68.6, 69.4, 65.3, 68.8

Using the Average, Standard Deviation function from Microsoft Excel, we have;

The mean for Catalyst 1, $\overline x_1$ = 65.22

The standard deviation for catalyst 1, σ₁ = 3.444416

The mean for Catalyst 2, $\overline x_2$ = 68.42

The standard deviation for catalyst 2, σ₂ = 2.22401

The 95% confidence interval on the difference in mean is given as follows;

$\left (\bar{x}_1-\bar{x}_{2} \right ) \pm z_{c}\sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}} + \dfrac{\sigma_{2}^{2}}{n_{2}}}$

The critical-z for a 95% confidence interval = 1.96

Therefore, we have;

$\left (65.22-68.42 \right ) \pm 1.96 \times \sqrt{\dfrac{3^{2}}{10} + \dfrac{3^{2}}{10}}$

Therefore, we have;

The 95% C.I. is approximately -5.83 < μ₁ - μ₂ < -0.57

The test statistics is given as follows;

$z=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}}$

Therefore we have;

$z=\dfrac{(65.22-68.42)}{\sqrt{\dfrac{3^{2} }{10}+\dfrac{3^{2}}{10}}} = -2.385$

The p-value = P(Z<-2.39 or Z > 2.39) = 2P(Z<-2.39) = 2 × 0.00842 = 0.01684

b. From the confidence interval which range from approximately -5.83 to -0.57 and does not include 0, therefore, there is a difference in mean active concentration which depends on the choice of catalyst

c. The power of the test

The sample mean difference is given as follows;

$\left (\bar{x}_1-\bar{x}_{2} \right ) = \pm z_{c}\sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}} + \dfrac{\sigma_{2}^{2}}{n_{2}}}$

Therefore, we have;

$\left (\bar{x}_1-\bar{x}_{2} \right ) = \pm 1.96 \times \sqrt{\dfrac{3^{2}}{10} + \dfrac{3^{2}}{10}} = \pm 2.6296$

The z-value is given as follows;

$z=\dfrac{(\bar{x}_{1}-\bar{x}_{2})-(\mu_{1}-\mu _{2} )}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}}$

μ₁ - μ₂ = 5

When $\bar{x}_1-\bar{x}_{2} = 2.6296$

$z=\dfrac{2.6296-5}{\sqrt{\dfrac{3^{2} }{10}-\dfrac{3^{2}}{10}}} \approx -1.7667918$

When $\bar{x}_1-\bar{x}_{2} = -2.6296$

$z=\dfrac{-2.6296-5}{\sqrt{\dfrac{3^{2} }{10}-\dfrac{3^{2}}{10}}} \approx -5.686768$

The power of the test is given by P = P(Z<-5.69) + P(Z>-1.77) = 0 + 0.03836 = 0.03836

The power of the test = 0.03836

d. The sample size is statistically adequate because the confidence interval of -5.83 < μ₁ - μ₂ < -0.57 has a value of -5 as a possible population difference in mean

The assumption of normality seems adequate because the confidence interval obtained by using the sample standard deviation is given as follows;

(-5.74, -0.66) which also contains -5 which is the difference in the population mean

4 0

3 years ago

Identify the vertex and directrix of each. y = x2 + 2x - 4

sergey [27]

Answer:

See below

Step-by-step explanation:

Convert equation to vertex form $y=(x-h)^2+k$ by completing the square: