I believe the answer is 2 3/7 I hope I could help
Answer:
(29.46 mm, 29.54 mm).
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution 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 = 43 - 1 = 42
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 42 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.018
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 29.5 mm - 0.04mm = 29.46 mm
The upper end of the interval is the sample mean added to M. So it is 29.5 mm + 0.04mm = 29.54 mm
The 95% confidence level for the true mean width is: (29.46 mm, 29.54 mm).
First, distribute 2 to x and 4, and -3 to x and -5 (distributive property)
2(x + 4) = 2x + 8
-3(x - 5) = -3x + 15
2x + 8 - 3x + 15
Next, combine like terms
2x - 3x = -x
8 + 15 = 23
-x + 23 is your answer
hope this helps
Answer: the bridge is 11.025 m high
Step-by-step explanation:
Given that;
Time taken to hit the water t = 1.5 sec
height of bridge = ?
lets take a look at the equation of motion;
y(t) = y₀ + (1/2)at²
with initial velocity zero and we know that acceleration due to gravidity is 9.8m/s
we substitute
y(t) = (1/2)gt² = (1/2) × 9.8 × (1.5)²
y(t) = (1/2)gt² = (1/2) × 9.8 × (1.5)²
y(1.5) = 11.025 m
Therefore, the bridge is 11.025 m high