The angle measure (120°) is the average of the measures of the two arcs (x, 2x). Hence
.. (x +2x)/2 = 120°
.. 3x = 240°
.. x = 80°
Answer:
50%
Step-by-step explanation:
12500 - 7500 = 5000
5000/100 = 50%
Well formatted version of the question can be found in the picture attached below :
Answer:
Both Carlos and Irene
Step-by-step explanation:
Given the expression (28x - 8)
Carlos : 2(14x - 4)
Danny : 7(4x - 1)
Irene : 4(7x - 2)
The factored expression Given by Carlos, Irene and Danny can be expanded to check if it gives the same expression as that Given in the question :
Carlos :
2(14x - 4)
2*14x - 2*4
28x - 8
Danny :
7(4x - 1)
7*4x - 7*1
28x - 7
Irene :
4(7x - 2)
4*7x - 4*2
28x - 8
From.the expanded solutions obtained ;
Both Carlos and Irene's solution corresponds to the factored form of the original equation and are Hence correct
1. x intercept is (10,0)
y intercept is (0,6)
2. x intercept is (4,0)
y intercept is (0, -1)
3. x intercept is ( 4,0)
y intercept is (0,8)
hope this helps :) !
Answer:
The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.
Step-by-step explanation:
Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.
The random variable <em>X</em> is exponentially distributed with mean 7 minutes.
Then the parameter of the distribution is,
.
The probability density function of <em>X</em> is:

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

![=\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148](https://tex.z-dn.net/?f=%3D%5Cint%5Climits%5E%7B9%7D_%7B6%7D%20%7B%5Cfrac%7B1%7D%7B7%7D%5Ccdot%20e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B7%7D%5Ccdot%20%5Cint%5Climits%5E%7B9%7D_%7B6%7D%20%7Be%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5B-e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%5D%5E%7B9%7D_%7B6%7D%5C%5C%5C%5C%3De%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%206%7D-e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%209%7D%5C%5C%5C%5C%3D0.424373-0.276453%5C%5C%5C%5C%3D0.14792%5C%5C%5C%5C%5Capprox%200.148)
Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.