Answer:
1.4in
Step-by-step explanation:
Length of Photo = 4in
Width of Photo = 3in
Unknown:
Value of X = ?
Solution:
Follow these steps:
Area of a rectangle = l x w
Since the photo is a rectangle; area of photo:
Area of photo = 4in x 3in = 12in²
For the area of the ad;
Length of ad = 4 + x
Width of ad = 3 + x
Given that,
the area of the photo =
area of ad
12in² =
area of ad
Area of ad = 24in²
Area of the ad;
(4 + x) (3 + x) = 24
12 + 4x + 3x + x² = 24
12 + 7x + x² = 24
x² + 7x = 24 - 12
x² + 7x = 12
x² + 7x - 12 = 0
Using the almighty formula where
a = 1, b = 7 and c = -12
x = 
x =
or ![\frac{-7 - \sqrt[]{-7^{2} - 4x1x-12 } }{2x1}](https://tex.z-dn.net/?f=%5Cfrac%7B-7%20-%20%5Csqrt%5B%5D%7B-7%5E%7B2%7D%20-%204x1x-12%20%7D%20%7D%7B2x1%7D)
x = 1.4 or -8.4
therefore the answer is 1.4in
x is 1.4in
In the first case we'd subtract 1 from both sides, obtaining |x-1|<14.
In the second case we'd also subtract 1 from both sides, and would obtain
|x-1|>14.
What would the graphs look like?
In the first case, the graph would be on the x-axis with "center" at x=1. From this center count 14 units to the right, and then place a circle around that location (which would be at x=15). Next, count 14 units to the left of this center, and place a circle around that location (which would be -13). Draw a line segment connecting the two circles. Notice that all of the solutions are between -13 and +15, not including these endpoints.
In the second case, x has to be greater than 15 or less than -13. Draw an arrow from x=1 to the left, and then draw a separate arrow from 15 to the right. None of the values in between are solutions.
Answer:
about 11.03 million
Step-by-step explanation:
Use the equation I = P(1+r/100)^n - P (I is the compound interest, P is the principle, r is the rate percent, and n is the number of years):
Substitute the values given:
I = 70,000,000(1 + 5/100)^3 - 70,000,000
Use a calculator to solve and you will get ~11.03 million.
Answer:
95% of the text messages have length between 23 units and 47 units.
Step-by-step explanation:
We are given the following in the question:
The lengths of text messages are normally distributed.
95% confidence interval:
(23,47)
Thus, we could interpret the confidence interval as:
About 95% of the text messages have length between 23 units and 47 units.
By Empirical rule for a normally distributed data, about 95% of data lies within 2 standard deviations of mean , thus we can write:

Thus, the mean length of text messages is 23 units and standard deviation is 6 units.