Which student expression is equal to 8×5

Answer:The following data gives the scores of 13 students in a Biostatistics exam. 75 80 28 70 95 82 75 64 61 90 81 65 91 a) Find the following statistical measures 1. Mean 2. Median 3. Mode 4. Range 5. 34th percentile 6. Interquartile Range 7. Variance 8. Standard deviation PRINCIPLES OF STATISTICS Assignment (1) Due Date: 15/7/2020 9. Coefficient of variation. b) (Without Calculations) If the instructor decide to add up 5 marks for every student, what are the values of the statistical measures mentioned in part (a). c) Construct the Boxplot of students' scores, and identify any possible outliers.

6 0

2 years ago

8(s+3)=72 i need help it is due tomorrow

Montano1993 [528]

Answer:

s = 6

Step-by-step explanation:

8(s+3) = 72

Expand the brackets out:

(8 x s) + (8 x 3) = 72

8s + 24 = 72

Now you get rid of 24 on the left, by taking away 24. But you also have to do the same on the right side so that both sides are equal:

8s + 24 = 72

left(-24) = right (-24)

So that leaves you with:

8s = 48

Now divide both sides by 8 to make them equal and to get s on its own:

8s = 72

left(/8) = right (/8)

Therfore:

s = 6

3 0

2 years ago

The following data on average daily hotel room rate and amount spent on entertainment (The Wall Street Journal, August 18, 2011)

Zielflug [23.3K]

Answer:

a. Predicted Amount = $109.46

b. Confidence Interval = (94.84,124.08)

c. Interval = (110.6883,188.8517)

Step-by-step explanation:

Given

ŷ = 17.49 + 1.0334x.

SSE = 1541.4

a.

ŷ = 17.49 + 1.0334(89)

ŷ = 109.4626

ŷ = 109.46 --- Approximated

Predicted Amount = $109.46

b.

ŷ = 17.49 + 1.0334(89)

ŷ = 109.4626

ŷ = 109.46

First we calculate the standard deviation

variance = SSE/(n-2)

v = 1541.4/(9-2)

v = 1541.4/7

v = 220.2

s = √v

s = √220.2

s = 14.839

Then we calculate mean(x) and ∑(x - (mean(x))²

X --- Y -- Mean(x) --- ∑(x - (mean(x))²

148 -- 161 -- 43-- 1849

96 || 105|| -9 || 81

91 ||101 || -14 || 196

110 || 142 || 5 || 25

90 || 100 || -15 || 225

102 || ||120 ||-3|| 9

136 || 167 ||31 ||961

90 || 140 ||-15 ||225

82 || 98 ||-23 || 529

Sum 945 || 1134|| 0 ||4100

Mean (x) = 945/9 = 105

∑(x - (mean(x))² = 4100

α = 1 - 95% = 5%

α/2 = 2.5% = 0.025

tα,df = n − 2 = t0.025,7 =2.365

Confidence interval = 109.46 ± 2.365 * 14.839 √((1/9)+ (89-105)²/4100

Confidence Interval = (109.46 ± 14.62)

Confidence Interval = (94.84,124.08)

c.

ŷ = 17.49 + 1.0334(128)

ŷ = 149.7652

ŷ = 149.77

Interval = 149.77 ± 2.365 * 14.839 √((1/9)+ (128-105)²/4100

Interval = 149.77 ± 39.0817

Interval = (110.6883,188.8517)

3 0

3 years ago