<span>4x +5=9x-2
9x - 4x = 5 + 2
5x = 7
x = 7/5
x = 1.4</span>
Using the <em>system of equation</em> created, Emily will catch up Lucy after 30 seconds
Given the Parameters :
- Lucy's distance = 2t
- Emily's distance = 5t
<u>We can set up an equation to represent the required scenario thus</u> :
Emily's distance = Lucy's distance + 90
5t = 2t + 90
We solve for t
<em>Collect like terms</em> :
5t - 2t = 90
3t = 90
Divide both sides by 3 to isolate t
t = 90/3
t = 30
Therefore, Emily will catch up with Lucy after 30 seconds
Learn more :brainly.com/question/13218948
Answer:
The interval [32.6 cm, 45.8 cm]
Step-by-step explanation:
According with the <em>68–95–99.7 rule for the Normal distribution:</em> If 
is the mean of the distribution and s the standard deviation, around 68% of the data must fall in the interval
![\large [\bar x - s, \bar x +s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-%20s%2C%20%5Cbar%20x%20%2Bs%5D)
around 95% of the data must fall in the interval
![\large [\bar x -2s, \bar x +2s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-2s%2C%20%5Cbar%20x%20%2B2s%5D)
around 99.7% of the data must fall in the interval
![\large [\bar x -3s, \bar x +3s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-3s%2C%20%5Cbar%20x%20%2B3s%5D)
So, the range of lengths that covers almost all the data (99.7%) is the interval
[39.2 - 3*2.2, 39.2 + 3*2.2] = [32.6, 45.8]
<em>This means that if we measure the upper arm length of a male over 20 years old in the United States, the probability that the length is between 32.6 cm and 45.8 cm is 99.7%</em>
Answer:
We accept H₀ with the information we have, we can say level of ozone is under the major limit
Step-by-step explanation:
Normal Distribution
population mean = μ₀ = 7.5 ppm
Sample size n = 16 df = n - 1 df = 15
Sample mean = μ = 7.8 ppm
Sample standard deviation = s = 0.8
We want to find out if ozono level, is above normal level that is bigger than 7.5
1.- Hypothesis Test
null hypothesis H₀ μ₀ = 7.5
alternative hypothesis Hₐ μ₀ > 7.5
2.-Significance level α = 0.01 we will develop one tail-test (right)
then for df = 15 and α = 0,01 from t -student table we get
t(c) = 2.624
3.-Compute t(s)
t(s) = ( μ - μ₀ ) / s /√n ⇒ t(s) = ( 7.8 - 7.5 )*4/0.8
t(s) = 0.3*4/0.8
t(s) = 1.5
4.-Compare t(s) and t(c)
t(s) < t(c) 1.5 < 2.64
Then t(s) is inside the acceptance region. We accept H₀
Answer:
Step-by-step explanation:
Let width = w units
Length = 4w - 3
Cost for lining corners of wall = $ 18.50
Perimeter of the wall = 18.50/0.25 = 74 ft
2*(length +width ) = 74
2*(4w - 3 + w) =74
2*(5w - 3) =74
10w - 6 = 74 {Add 6 to both sides}
10w - 6 + 6 = 74 + 6
10w = 80
Divide both sides by 10
10w/10 = 80/10
w = 8 ft
Length = 4w - 3 = 4*8 - 3 = 32 - 3
Length = 29 ft
Area of the wall = length * width
= 29 * 8
= 232 square ft
Cost of painting the wall per square feet = 20 cents = 0.20
Cost of painting 232 square feet = 232 * 0.20
= $ 46.40
