How am I suppose to answer this question

Eight trials are simulated. The results are shown in the table.

bezimeni [28]

Answer:

2.897

Step-by-step explanation:

Let us find the mean and variance for the sample first.

105 104 110 112

114 106 108 109

Mean = sum/8 = 108.5

Variance = 12

Std dev = sq rt of variance = 3.464

Std error = std dev/ sq rt n

since n =8, we get std error = 1.225

Since sample size is small, df =8-1 =7

For 95% confidence intervals, t critical value for two tailed=2.365

Margin of error = std error x t critical = 1.225(2.365)

=2.897

3 0

2 years ago

Derive Euler's formula e^(ix) - cos(x) + i*sin(x) by using taylor series expansion

gavmur [86]

Answer:

the answer is probably 5

5 0

2 years ago

For which pair of functions is the vertex of k(x)7 units below the vertex of f(x)?

kvv77 [185]

Answer: Option C

$f(x) = x^2;\ k (x) = x ^ 2 -7$

Step-by-step explanation:

Whenever we have a main function f(x) and we want to transform the graph of f(x) by moving it vertically then we apply the transformation:

$k (x) = f (x) + b$

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units down.

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units upwards.

In this case we have

$f (x) = x ^ 2$

We know that this function has its vertex in point (0,0).

Then, to move its vertex 7 units down we apply the transformation:

$k (x) = f (x) - 7\\\\k (x) = x ^ 2 -7$ .

Then the function k(x) that will have its vertex 7 units below f(x) is

$k (x) = x ^ 2 -7$

3 0

2 years ago

If sinA+cosecA=3 find the value of sin2A+cosec2A

Irina18 [472]

Answer:

$\sin 2A + \csc 2A = 2.122$

Step-by-step explanation:

Let $f(A) = \sin A + \csc A$ , we proceed to transform the expression into an equivalent form of sines and cosines by means of the following trigonometrical identity:

$\csc A = \frac{1}{\sin A}$ (1)

$\sin^{2}A +\cos^{2}A = 1$ (2)

Now we perform the operations: $f(A) = 3$

$\sin A + \csc A = 3$

$\sin A + \frac{1}{\sin A} = 3$

$\sin ^{2}A + 1 = 3\cdot \sin A$

$\sin^{2}A -3\cdot \sin A +1 = 0$ (3)

By the quadratic formula, we find the following solutions:

$\sin A_{1} \approx 2.618$ and $\sin A_{2} \approx 0.382$

Since sine is a bounded function between -1 and 1, the only solution that is mathematically reasonable is:

$\sin A \approx 0.382$

By means of inverse trigonometrical function, we get the value associate of the function in sexagesimal degrees:

$A \approx 22.457^{\circ}$

Then, the values of the cosine associated with that angle is:

$\cos A \approx 0.924$

Now, we have that $f(A) = \sin 2A +\csc2A$ , we proceed to transform the expression into an equivalent form with sines and cosines. The following trignometrical identities are used: