Simplify both sides of the equation
-25(x-4)=-55(x-10)
(-25)(x) + (-25)(-4)=(-55)(x) + (-55) (-10)
Distribute
-25x+100=-55x+550
Add 55x to both sides
-25x+100+55x=-55x+550+55x
30x+100-100=550-100
subtract 100 from both sides
30x+100-100=550-100
30x=450
divide both sides by 30
30x/30=450/30
x= 15
The value of x is 15
I hope that's help you my friend !
Answer:
2√15
Step-by-step explanation:
Use the Pythagorean theorem.
2² + x² = 8²
x² + 4 = 64
x² = 60
x² = 4 * 15
x = 2√15
Answer:
C. x^14
Step-by-step explanation:
(x^9)^0 is 1
(x^7)^2 is x^14
x^14*1 is x^14
The Answer is C. x^14
Hope this helps!
Answer:
Probability of stopping the machine when
is 0.0002
Probability of stopping the machine when
is 0.0013
Probability of stopping the machine when
is 0.0082
Probability of stopping the machine when
is 0.0399
Step-by-step explanation:
There is a random binomial variable
that represents the number of units come off the line within product specifications in a review of
Bernoulli-type trials with probability of success
. Therefore, the model is
. So:
![P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%209%29%20%3D%201%20-%20P%20%28X%20%5Cgeq%209%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%209%7D%20%280.91%29%5E%7B9%7D%280.09%29%5E%7B6%7D%2B...%2B%7B%2015%20%5Cchoose%2015%7D%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0002%20)
![P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%2010%29%20%3D%201%20-%20P%20%28X%20%5Cgeq%2010%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%2010%7D%280.91%29%5E%7B10%7D%280.09%29%5E%7B5%7D%2B...%2B%7B15%20%5Cchoose%2015%7D%20%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0013%20)
![P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%2011%29%20%3D%201%20-%20P%20%28X%20%5Cgeq%2011%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%2011%7D%280.91%29%5E%7B11%7D%280.09%29%5E%7B4%7D%2B...%2B%7B15%20%5Cchoose%2015%7D%20%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0082)
![P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%2012%29%20%3D%201-%20P%20%28X%20%5Cgeq%2012%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%2012%7D%280.91%29%5E%7B12%7D%280.09%29%5E%7B3%7D%2B...%2B%7B15%20%5Cchoose%2015%7D%20%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0399%20)
Probability of stopping the machine when
is 0.0002
Probability of stopping the machine when
is 0.0013
Probability of stopping the machine when
is 0.0082
Probability of stopping the machine when
is 0.0399
Answer:
$9000 at 4$
and
$10000 at 8%
Step-by-step explanation:
Let's assume that "x" is the amount deposited in the 4% account and "y" is the amount deposited in the 8% account.
Recall the formula for interest as : 
where I is the interest, R is the annual rate of interest and t is the number of years.
Since there are two investments, we need to add both interests at the end of the one year: I1 = x (0.04) (1) = 0.04 x and I2 = y (0.08) (1) = 0.08 y
Total Interest = Interest (from the 4% account) + Interest (from the 8% account)
Total Interest = $1160 = 0.04 x + 0.08 y
we also know that the total invested (x + y) adds to $19,000, that is:
$19,000 = x + y
Then we can solve these system of two equations by substitution, for example solving for y in the second equation and using the y substitution in the first equation;
y = 19000 - x
1160 = 0.04 x + 0.08 (19000 - x)
1160 = 0.04 x + 1520 - 0.08 x
0.08 x - 0.04 x = 1520 - 1160
0.04 x = 360
x = 360/0.04 = $9000
Then the other investment was : y = $19000 - $9000 = $10000