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Solnce55 [7]
3 years ago
10

PLEASE HELP!!!!! The table displays the mean name length for seven samples of students. Sample Mean Name Length 1 5.4 2 7.1 3 6.

3 4 5.2 5 6.0 6 4.9 7 6.2 What can be said about the variation between the sample means? A. The variation between the sample means is small. B. The variation between the sample means is large. C. The variation shows that the values are far apart. D. The variation cannot be used to make predictions.
Mathematics
2 answers:
Vikki [24]3 years ago
8 0

Answer:

A. The variation between the sample means is small.

Step-by-step explanation:

I took the quiz and got it right. O gotchu fam

kolezko [41]3 years ago
5 0

Answer:

I think its A!

Step-by-step explanation:

Im currently taking the quiz right now but ill edit this if i get it wrong!

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Convert the credit card rate to the APR. Ohio, 0.02192% daily rate. Please round your answer to the nearest percent. (zero decim
kogti [31]

Answer:

8%

Step-by-step explanation:

APR means annual percentage rate.

To convert the daily rate to an APR, multiply the daily rate by the number of days in a year

365 days = 1 year

0.02192% x 365 = 8.0008%

To round off to the nearest percent, look at the first number after the decimal, if it is less than 5, add zero to the units term, If it is equal or greater than 5, add 1 to the units term.

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3 0
3 years ago
Question 21 pts
Pachacha [2.7K]

Answer:

x = -2 or x = -6 or x = sqrt(5) or x = -sqrt(5)

Step-by-step explanation:

Solve for x:

x^4 + 8 x^3 + 7 x^2 - 40 x - 60 = 0

The left hand side factors into a product with three terms:

(x + 2) (x + 6) (x^2 - 5) = 0

Split into three equations:

x + 2 = 0 or x + 6 = 0 or x^2 - 5 = 0

Subtract 2 from both sides:

x = -2 or x + 6 = 0 or x^2 - 5 = 0

Subtract 6 from both sides:

x = -2 or x = -6 or x^2 - 5 = 0

Add 5 to both sides:

x = -2 or x = -6 or x^2 = 5

Take the square root of both sides:

Answer: x = -2 or x = -6 or x = sqrt(5) or x = -sqrt(5)

8 0
3 years ago
PLEASE HELP.!! THANK YOUU. accurate answers appreciated:)
Anni [7]

\bf \cfrac{1}{1-sin(x)}+\cfrac{1}{1+sin(x)}=\cfrac{2}{cos^2(x)} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the LCD of [1-sin(x)][1+sin(x)]}}{\cfrac{[1+sin(x)]1~~+~~[1-sin(x)]1}{\underset{\textit{difference of squares}}{[1-sin(x)][1+sin(x)]}}} \\\\\\ \cfrac{1+sin(x)+1-sin(x)}{1^2-sin^2(x)}\implies \cfrac{1+sin(x)+1-sin(x)}{1-sin^2(x)}

recall that 1 - sin²(θ) = cos²(θ).

5 0
3 years ago
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