In the diagram shown, the distance between points A and C is the same as the distance between points B and G.

pepper jackie decides her map of the youth center is too small. She draws a new map of the youth center that is 3 times larger.

Kitty [74]

Solution:

Let starting map of youth center has Dimension (Preimage)

1 unit of y= k units

1 unit of x = t units

The new map of youth center which is 3 times larger than original(Image)

1 unit of y= 3 k units

1 unit of x = 3 t units

Image = 3 × Pre- image

The image is larger than Pre- image.

So, Scale of the new map = 3 × The original dimension of map.

7 0

3 years ago

Read 2 more answers

It is known that 10% of the calculators shipped from a particular factory are defective. What is the probability that no more th

ycow [4]

Answer: 0.9477

Step-by-step explanation:

p = 10% = 0.1, q = 90% = 0.9, n = 4

The question follows a binomial probability distribution since the experiment (defectiveness of random sample of calculator) is performed more than once ( 4 calculators are defective).

The question is to find the probability that not more than one in a random sample of 4 calculators is defective that's

p(x≤1) = p(x=0) + p(x=1)

The probability mass function of a binomial probability distribution is given below as

P(x=r) =nCr × p^r × q^n-r

At x = 0

p(x=0) = 4C0 × 0.1^0 × 0.9^4-0

p(x=0) = 4C0 × 0.1^0 × 0.9^4

p(x=0) = 1 × 1 × 0.6561

p(x=0) = 0.6561.

At x = 1

p(x=1) = 4C1 × 0.1^1 × 0.9^4-1

p(x=1) = 4C1 × 0.1^1 × 0.9^3

p(x=1) = 4 × 0.1 × 0.729

p(x=1) = 0.2916

p(x≤1) = 0.6561 + 0.2916 = 0.9477

7 0

4 years ago

A linear function has a slope of 5/2 and passes through the point 0,6 . What is the equation of the line?

a_sh-v [17]

Answer:

y = x^ 3 - 17 x ^2 /2 + 15 x

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The students in Mr. Kim's class are discussing how to apply the properties of equality. Jared started with the equation 10 - 4=

iVinArrow [24]

<u>Answer</u>:

Jared is right.

<u>Explanation</u>:

proof 1:

10 - 4 = 6

6 = 6

second proof:

10-3 - 4 = 6 - 3

10 - 7 = 3

3 = 3

third proof:

10 - 4 - 3 = 6 - 3

10 - 7 = 3

3 = 3

Hence for all LHS = RHS proved and Jared is correct.

7 0

3 years ago

Grades on a standardized test are known to have a mean of 1000 for students in the United States. The test is administered to 45

vovikov84 [41]

Answer:

a. The 95% confidence interval is 1,022.94559 < μ < 1,003.0544

b. There is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. i. The 95% confidence interval for the change in average test score is; -18.955390 < μ₁ - μ₂ < 6.955390

ii. There are no statistical significant evidence that the prep course helped

d. i. The 95% confidence interval for the change in average test scores is 3.47467 < μ₁ - μ₂ < 14.52533

ii. There is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

Step-by-step explanation:

The mean of the standardized test = 1,000

The number of students test to which the test is administered = 453 students

The mean score of the sample of students, $\bar{x}$ = 1013

The standard deviation of the sample, s = 108

a. The 95% confidence interval is given as follows;

$CI=\bar{x}\pm z\dfrac{s}{\sqrt{n}}$

At 95% confidence level, z = 1.96, therefore, we have;

$CI=1013\pm 1.96 \times \dfrac{108}{\sqrt{453}}$

Therefore, we have;

1,022.94559 < μ < 1,003.0544

b. From the 95% confidence interval of the mean, there is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. The parameters of the students taking the test are;

The number of students, n = 503

The number of hours preparation the students are given, t = 3 hours

The average test score of the student, $\bar{x}$ = 1019

The number of test scores of the student, s = 95

At 95% confidence level, z = 1.96, therefore, we have;

The confidence interval, C.I., for the difference in mean is given as follows;

$C.I. = \left (\bar{x}_{1}- \bar{x}_{2} \right )\pm z_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

Therefore, we have;

$C.I. = \left (1013- 1019 \right )\pm 1.96 \times \sqrt{\dfrac{108^{2}}{453}+\dfrac{95^{2}}{503}}$

Which gives;

-18.955390 < μ₁ - μ₂ < 6.955390

ii. Given that one of the limit is negative while the other is positive, there are no statistical significant evidence that the prep course helped

d. The given parameters are;

The number of students taking the test = The original 453 students

The average change in the test scores, $\bar{x}_{1}- \bar{x}_{2}$ = 9 points

The standard deviation of the change, Δs = 60 points

Therefore, we have;

C.I. = $\bar{x}_{1}- \bar{x}_{2}$ + 1.96 × Δs/√n

∴ C.I. = 9 ± 1.96 × 60/√(453)

i. The 95% confidence interval, C.I. = 3.47467 < μ₁ - μ₂ < 14.52533

ii. Given that both values, the minimum and the maximum limit are positive, therefore, there is no zero (0) within the confidence interval of the difference in of the means of the results therefore, there is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

5 0

3 years ago