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3 years ago

J and k stands for two numbers.

Mathematics

1 answer:

Katena32 [7]3 years ago

8 0

2j = .5k

Let j = 5
2(5) = .5k
10 = .5k
divide both sides by .5
20 = k
When j = 5 , k = 20

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Does the frequency distribution appear to have a normal distribution using a strict interpretation of the relevant criteria?Tem

Xelga [282]

Answer:

B. No, this distribution does not appear to be normal

Step-by-step explanation:

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3 years ago

Use the given sample data to construct the indicated confidence interval for the population mean. The principal randomly selecte

Lunna [17]

Answer:

The 90% confidence interval

(74.71, 82.63)

Step-by-step explanation:

Confidence Interval Formula is given as:

Confidence Interval = μ ± z (σ/√n)

Where

μ = mean score

z = z score

N = number of the population

σ = standard deviation

The mean is calculated as = The average of their scores

N = 6 students

(71.6 + 81.0 + 88.9 + 80.4 + 78.1 + 72.0 )/ 6

Mean score = 472/6

= 78.666666667

≈ 78.67

We are given a confidence interval of 90% therefore the

z score = 1.645

Standard Deviation for the scores =

s=(x -σ)²/ n - 1 =(71.6 - 78.67)²+(81.0 - 78.67)²+(88.9 - 78.67)² + (80.4 - 78.67)²+ (78.1 - 78.67)²+( 72.0 - 78.67)2/ 6 - 1

= 5.886047531

= 5.89

The confidence interval is calculated as

= μ ± z (σ/√N)

= 78.67 ± 1.645(5.89/√6)

= 78.67 ± 3.9555380987

The 90% confidence interval

is :

78.67 + 3.9555380987 = 82.625538099

78.67 - 3.9555380987 = 74.714619013

Therefore, the confidence interval is approximately between

(74.71, 82.63)

3 0

3 years ago

Find h(-1) if h(x) =- 3|3x – 6|– 3?

Dahasolnce [82]

Answer:

Step-by-step explanation:

h(-1) = - 3|3(-1) - 6| - 3

h(-1) = - 3|-3 - 6| - 3

h(-1) = - 3|-9| - 3

h(-1) = - 3(9) - 3

h(-1) = - 27 - 3

h(-1) = - 30

6 0

3 years ago

A certain test preparation course is designed to help students improve their scores on the GMAT exam. A mock exam is given at th

Bogdan [553]

Answer:

95% confidence interval for the average net change is between a lower limit of 11.709 and an upper limit of 30.921.

The critical value that should be used in constructing the confidence interval is 4.303.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = (17+20+26)/3 = 63/3 = 21

sd = sqrt[((17-21)^2 + (20-21)^2 + (26-21)^2) ÷ 3] = sqrt(42÷3) = sqrt(14) = 3.74

n = 3

degree of freedom = n-1 = 3-1 = 2

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value (t) corresponding to 2 degrees of freedom and 5% significance level is 4.303.

The critical value is 4.303

E = t×sd/√n = 4.303×3.74/√3 = 9.291

Lower limit of mean = mean - E = 21 - 9.291 = 11.709

Upper limit of mean = mean + E = 21 + 9.291 = 30.291

95% confidence interval is (11.709, 30.391)

8 0

3 years ago

What is the volume of the cone?

zhuklara [117]

Answer:

Step-by-step explanation:

d=22 yd

r=11 yd

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3 0

3 years ago