In the adjoining figure, PQRS is a square and TQRS is a trapezium. If the area of the square PQRS is 25sq.cm and TP=12cm, find t
he area of trapezium TQRS.
1 answer:
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In a trigonometric function
F(X) = ±UD ± A.COS(Px + LR)
UD - move the graph to Up or Down (+ = up | - = down)
A - amplitude
P - period (period = 2π/P)
LR - move the graph to Left or Right (+ = left | - = right)
So:
A) F(X) = COS(X + 1)
standard cosine graph with 1 unit shift to the left
B) F(X) = COS(X) - 1 = -1 + COS(X)
standard cosine graph with 1 unit down
C) F(X) = COS(X - 1)
standard cosine graph with shift 1 unit to the right
D) F(X) = SEN(X - 1)
standard Sine graph with shift 1 unit to the right
Observing the graph we notice the sine function shifted 1 unit to the right, then
<h3>Option D</h3><em>(cosine star the curve in X and Y = zero. sine start the curve in Y = 1)</em>
Answer:
second one and I don't know how to do it sorry is it a
Step-by-step explanation:
can you help me with my question
Answer:
The critical value is T = 1.895.
The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution.
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 = 6
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 6 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.895, which is the critical value.
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 60.46 - 12.301 = $48.159
The upper end of the interval is the sample mean added to M. So it is 60.46 + 12.301 = $72.761
The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761