When does the confirmation bias occur?

maxonik [38]

Answer:

a) well we know the hypotenuse and we know that all sides are 90 degrees, and that its an even square, so if we split it into a triangle , we know that the other 2 angles are 45 degrees total, with this we can calculate the other 2 sides, which round to 10 inches.

b) if we know the sides of the first side, knowing the fact that its a square we know that all sides are even, therefore, all sides are 10 inches, the formula the finding the perimeter is basic:

P=L(2) + W(2)

L being length, W being width and P being perimeter

so we add up all 4 sides, here is the equation:

10 + 10 + 10 + 10 = 10 x 4 = 40

With this, we know that the perimeter is 40 inches total

c) knowing how long the sides are, we can figure out the area through another basic formula:

A=L X W

With A meaning Area

since its a square, we know all the sides are even, so the width and length are both 10...

10 X 10 = 100

Step-by-step explanation:

I Hope I Helped!

3 0

3 years ago

The equation that models the current water temperature t of the swimming pool is t -6=78 which best describes the error made whe

Serga [27]

78! is the answer to this question! trust! -6

3 0

2 years ago

377. Dawn wants to compare the volume of a basketball with the

olchik [2.2K]

Answer:

$V = \frac{4}{3} \pi R^{3}$ and

$v = \frac{4}{3} \pi r^{3}$

Step-by-step explanation:

Dawn wants to compare the volume of a basketball with the volume of a tennis ball.

Now, if the radius of the basketball is R and that of the tennis ball is r, then the formula that can be used to determine the volume of each ball will be

$V = \frac{4}{3} \pi R^{3}$ and

$v = \frac{4}{3} \pi r^{3}$

Now, dividing V by v we can compare the volume of the basketball and the tennis ball. (Answer)

3 0

3 years ago

Will improving customer service result in higher stock prices for the companies providing the better service? "When a company’s

Orlov [11]

Question:

Company 2007 Score 2008 Score

Rite Aid 73 76

Expedia 75 77

J.C. Penney 77 78

a. For Rite Aid, is the increase in the satisfaction score from 2007 to 2008 statistically significant? Use α= .05. What can you conclude?

b. Can you conclude that the 2008 score for Rite Aid is above the national average of 75.7? Use α= .05.

c. For Expedia, is the increase from 2007 to 2008 statistically significant? Use α= .05.

d. When conducting a hypothesis test with the values given for the standard deviation,

sample size, and α, how large must the increase from 2007 to 2008 be for it to be statistically significant?

e. Use the result of part (d) to state whether the increase for J.C. Penney from 2007 to 2008 is statistically significant.

Answer:

a. There is sufficient statistical evidence to suggest that the increase in satisfaction score for Rite Aid from 2007 to 2008 is statistically significant

b. There is sufficient statistical evidence to suggest that the 2008 Rite Aid score, is above the national average of 75.7

c. The statistical evidence support the claim of a significant increase from 2007 to 2008

d. 1.802 and above is significant

e. The increase of J. C. Penney from 2007 is not statistically significant.

Step-by-step explanation:

Here we have

n = 60

σ = 6

μ₁ = 73

μ₂ = 76

We put H₀ : μ₁ ≥ μ₂ and

Hₐ : μ₁ < μ₂

From which we have;

$z=\frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}+\frac{\sigma _{2}^{2}}{n_{2}}}} = \frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{2\sigma_{}^{2} }{n_{}}}}}$

Plugging in the values we have

$z = \frac{(73-76)}{\sqrt{\frac{2\times 6^{2} }{60_{}}}}} = -2.7386$

The probability, P from z function computation gives;

P(Z < -2.7386) = 0.0031

Where we have P < α, we reject the null hypothesis meaning that there is sufficient statistical evidence to suggest that the increase in satisfaction score for Rite Aid from 2007 to 2008 is statistically significant

b. To test here, we have

H₀ : μ ≤ 75.7

Hₐ : μ > 75.7

The test statistic is given as follows;

$z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}} = \frac{76-75.7 }{\frac{6 }{\sqrt{60}}} = 0.3873$

Therefore, we have the probability, P given as the value for the function at z = 0.3873 that is we have;

P = P(Z > 0.3873) = P(Z < -0.3873) = 0.3493

Therefore, since P > α which is 0.05, we fail to reject the null hypothesis, that is there is sufficient statistical evidence to suggest that the 2008 Rite Aid score, is above the national average of 75.7

c. Here we put

Null hypothesis H₀ : μ₁ ≥ μ₂

Alternative hypothesis Hₐ : μ₁ < μ₂

The test statistic is given by the following equation;

$z=\frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}+\frac{\sigma _{2}^{2}}{n_{2}}}} = \frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{2\sigma_{}^{2} }{n_{}}}}}$

Plugging in the values we have

$z = \frac{(75-77)}{\sqrt{\frac{2\times 6^{2} }{60_{}}}}} = -1.8257$

The probability, P from z function computation gives;

P(Z < -1.8257) = 0.03394

The statistical evidence support the claim of a significant increase

d. For statistical significance at 0.05 significant level, we have z = -1.644854

Therefore, from;

$z=\frac{(\bar{x_{1}}-\bar{x_{2}})-(\mu_{1}-\mu _{2} )}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}$ . we have;

$z \times \sqrt{\frac{\sigma_{1}^{2} }{n_{1}}+\frac{\sigma _{2}^{2}}{n_{2}}} + (\mu_{1}-\mu _{2} )}{} ={(\bar{x_{1}}-\bar{x_{2}})$

Which gives

${(\bar{x_{1}}-\bar{x_{2}}) = z \times \sqrt{\frac{2\sigma_{}^{2} }{n_{}}}} + (\mu_{1}-\mu _{2} )}{} = -1.644854 \times \sqrt{\frac{2\times 6_{}^{2} }{60_{}}}} + 0 = -1.802$

Therefore an increase of 1.802 and above is significant

e. Based on the result of part d. we have for J.C. Penney from 2007 to 2008 an increase of 1 which is less than 1.802 at 5% significant level, is not significant.

5 0

3 years ago

A box plot is also known as a(n) box-and-whisker plot. Use the box plot below to find each of the following.

VLD [36.1K]

Answer:

the min is 53 the max is 90 median is 65 first quartile is 60 and the third quartile is 82

Step-by-step explanation:

3 0

3 years ago