Answer:
a)2*(15-s) mi
b)30mi
c) 30 mi
Step-by-step explanation:
1.) 10² + 4 · (6 + 3) - (27 ÷ 9) - 6 = 127
For these types of questions, try to think of it like, well since the last number is negative 6, let me try to get everything before it to equal 133 so it will end up being 127. :)
2.)

÷

·(0.3 + 0.42)
= 1/10 ÷ 1/2 + 1.6 · (0.72)
= 1/5 + 1.152
= 1.352
Remember that dividing a fraction by a fraction is like multiplying the coefficient of the second one to the original one.
3.) 15 · 1/2 ÷ 3/4
Remember that product is multiplication and quotient is division.
4.)

Remember that difference is
subtraction and separating into three parts is like
dividing it into three parts.
Hope this helped and ask me for further explanation! ^-^
a. Find the probability that an individual distance is
greater than 214.30 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (214.30 – 205) / 8.3
z = 1.12
Since we are looking for x > 214.30 cm, we use the
right tailed test to find for P at z = 1.12 from the tables:
P = 0.1314
b. Find the probability that the mean for 20 randomly
selected distances is greater than 202.80 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (202.80 – 205) / 8.3
z = -0.265
Since we are looking for x > 202.80 cm, we use the
right tailed test to find for P at z = -0.265 from the tables:
P = 0.6045
c. Why can the normal distribution be used in part (b),
even though the sample size does not exceed 30?
I believe this is because we are given the population
standard deviation sigma rather than the sample standard deviation. So we can
use the z test.
The equation would be 2x+3 so when you plug it in it's 2(-2)+3 which simplify to -4+3 which is -1
Answer:
17.96m/s
Step-by-step explanation:
Based on law of conservation of momentum;
m1u1 +m2u2 =(m1+m2)v
v is the velocities of the car after impact
Substitute the given values
1300(20)+900(15) = (1300+900)v
26000 + 13500 = 2200v
39500 = 2200v
v = 39500/2200
v = 17.96m/s
Hence the speed of the cars after impact is 17.96m/s