The value of 4 less than the product of 0.25 and x is greater than 6. What are all the possible values of x? SHOW YOUR WORK!

a class conduct a survey of 1000 students 6% of the students have forgotten their locker combinations how many students have for

solong [7]

This is a simple percentage question. 6/100 * 1000 = 60 people have forgotten

3 0

2 years ago

today only, a suit is being sold at a 29% off (discount). The sale price is $284. What was the original price?

natta225 [31]

Answer:

201.64

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A rectangle with vertices located at (1,2), (1,5), (3,5), and (3,2) is stretched horizontally by a factor of 3 with respect to t

Vika [28.1K]

Answer:

18 square units

Step-by-step explanation:

The given rectangle has an x-dimension of 3-1 = 2, and a y-dimension of 5-2 = 3. Stretching the x-dimension by a factor of 3 will multiply it to ...

2×3 = 6

and give you a rectangle 6 units wide by 3 units high. Its area will be ...

(6 units)(3 units) = 18 units²

5 0

3 years ago

Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body b

kramer

Answer:

(a)

$\bar x = 0.0075$

$s = 0.0049$

(b)

B. The sample is too small to make judgments about skewness or symmetry.

Step-by-step explanation:

Given:

$n = 8$

$\begin{array}{ccccccccc}{} & {S} & {u} & {b} &{j} & {e} & {c} & {t} & {s} &{Operator} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {1} & 1.326 & 1.337 & 1.079 & 1.229 & 0.936 & 1.009 & 1.179 & 1.289 & 2 & 1.323 & 1.322 & 1.073 & 1.233 & 0.934 & 1.019 & 1.184 & 1.304 \ \end{array}$

Solving (a):

First, calculate the difference between the recorded TBBMC for both operators:

$\begin{array}{ccccccccc}{} & {S} & {u} & {b} &{j} & {e} & {c} & {t} & {s} &{Operator} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {1} & 1.326 & 1.337 & 1.079 & 1.229 & 0.936 & 1.009 & 1.179 & 1.289 & 2 & 1.323 & 1.322 & 1.073 & 1.233 & 0.934 & 1.019 & 1.184 & 1.304 &{|1 - 2|} &0.003 & 0.015 & 0.006 & 0.004 & 0.002 & 0.010 & 0.005 & 0.015 \ \end{array}$

The last row which represents the difference between 1 and 2 is calculated using absolute values. So, no negative entry is recorded.

The mean is then calculated as:

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{0.003 + 0.015 + 0.006 + 0.004 + 0.002 + 0.010 + 0.005 + 0.015}{8}$

$\bar x = \frac{0.06}{8}$

$\bar x = 0.0075$

Next, calculate the standard deviation (s).

This is calculated using:

$s = \sqrt{\frac{\sum (x - \bar x)^2}{n}}$

So, we have

$s = \sqrt\frac{(0.003 - 0.0075)^2 + (0.015 - 0.0075)^2+ (0.006- 0.0075)^2+ ....... + (0.005- 0.0075)^2+ (0.015- 0.0075)^2 }{8}$ $s = \sqrt\frac{0.00019}{8}$