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3 years ago

PLEASEEE SOMEONE HELP ME ON THIS!!!! It’s due in a couple hours

Mathematics

2 answers:

labwork [276]3 years ago

8 0

Answer:

Step-by-step explanation:

This to the nearst tenth in decimal

zimovet [89]3 years ago

4 0

Answer:

x = 8.0 m

Step-by-step explanation:

With the angle presented, I'm assuming you know basic trigonometry. We know cosθ = adj/hyp from SOH CAH TOA. We have angle and hypotenuse, so plugging in those values we get cos(20°) = x / 8.5. Solve for x to get 8.5cos(20°) = 7.987387277 m = 8.0 m.

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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

5 0

3 years ago

Im cyril anne bato you what your name

Andrej [43]

Answer:

heyyyyyyyyyyyyyyyyyyyyyyyyy

4 0

3 years ago

Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule.

vekshin1

Simplify:

A(n) = -6 + (n - 1)(6) = -6 + (n)(6) + (-1)(6) = -6 + 6n - 6 = 6n - 12

Put n = 1, n = 4 and n = 10 to the expression:

A(1) = 6(1) - 12 = 6 - 12 = -6

A(4) = 6(4) - 12 = 24 - 12 = 12

A(10) = 6(10) - 12 = 60 - 12 = 48

<h3>Answer:</h3><h3>first term = -6, fourth term = 12 and tenth term = 48.</h3><h3>-6, 12, 48</h3>

7 0

3 years ago

Read 2 more answers

Define an action adder :: io () that reads a given number of integers from the key board, one per line, and displays their sum.

aleksandrvk [35]

The program is an illustration of loops

<h3>What are loops?</h3>

Loops are program statements that are used to perform repetition

<h3>The main program</h3>

The program written in Python, where comments are used to explain each line is as follows:

#This initializes sum to 0

summ = 0

#This gets input for the first number

num = int(input())

#This is repeated while num is not -1

while num!= -1:

#This calculates the sum

summ+=num

#This gets input for the num

num = int(input())

#This prints the sum

print(summ)