Answer:
The value of the ve = 9m/sec
Step-by-step explanation:
From the given formula, it can be conclude that this is evenly accelerated movement.
First I will rewrite given formula
vi = √ ve∧2 - 2ad First we will square on both sides and get
vi∧2 = ve∧2 - 2ad Now we will add monom (+2ad) to both sides and get
vi∧2 + 2ad = ve∧2 -2ad + 2ad =>
ve∧2 = vi∧2 +2ad Now we will rooted both sides and get
ve = √vi∧2 + 2ad
Now we will replace given data vi=7m/sec, a=8m/s∧2 and d=2m in the last formula
ve= √7∧2 + 2*8*2 = √49+32 = √81 = 9
ve= 9m/sec
Good luck!!!
Answer: N >/= 7 bits
Minimum of 7 bits
Step-by-step explanation:
The minimum binary bits needed to represent 65 can be derived by converting 65 to binary numbers and counting the number of binary digits.
See conversation in the attachment.
65 = 1000001₂
65 = 7 bits :( 0 to 2^7 -1)
The number of binary digits is 7
N >/= 7 bits
Answer:

Step-by-step explanation:
(4/3)P -(4/3)A + A = B . . . . . . add A
(4P -A)/3 = B . . . . . . . . . . . . . simplify
Then the coordinates of point B are ...
B = (4(1, 6) -(-5, 3))/3 = (9, 21)/3
B = (3, 7)
Answer:
92%
Step-by-step explanation:
This is because all you have to do to turn a decimal into percentage is to move the decimal over 2 spots, and drop the Zero.
Answer:
The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 8 - 1 = 7
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 7 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.3246
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 0.32 - 0.0502 = 0.2698 mg/kg
The upper end of the interval is the sample mean added to M. So it is 0.32 + 0.0502 = 0.3702 mg/kg
The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).