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3 years ago

7. Find mDBC. 6. Find mŁABD.

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

8 0

ABD: 22
DBC: 109

they are congruent angles

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Service calls arriving at an electric company follow a Poisson distribution with an average arrival rate of 5656 per hour. Find

liberstina [14]

Answer:

The average number of service calls in a 15-minute period is of 14, with a standard deviation of 3.74.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval. The variance is the same as the mean.

Average rate of 56 calls per hour:

This means that $\mu = 56n$ , in which n is the number of hours.

Find the average and standard deviation of the number of service calls in a 15-minute period.

15 minute is one fourth of a hour, which means that $n = \frac{1}{4}$ . So

$\mu = 56n = \frac{56}{4} = 14$

The variance is also 14, which means that the standard deviation is $\sqrt{14} = 3.74$

The average number of service calls in a 15-minute period is of 14, with a standard deviation of 3.74.

4 0

3 years ago

Helllpppp!!!!!it is a log equation

Vaselesa [24]

I hope this is right

2log $2log6(2)+3log6(t)+1/2xlog6(y)$

6 0

2 years ago

For the function g(x) = x – 2x^2, find g (–3) + 13

svp [43]

Answer:

Step-by-step explanation:

g(x) = x - 2x²

To find g( - 3) + 13 substitute the value of x that's - 3 into g(x) and add 13 to it

We have

g( - 3) + 13 = - 3 - 2(-3)² + 13

= -- 3 - 2(9) + 13

= 3 - 18 + 13

We have the final answer as

Hope this helps you

8 0

3 years ago

Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|

Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0

3 years ago

The solution of the equation 2d - 6 = 5 is _____.

xeze [42]

Answer: d = 11/2 or 5 1/2

Step-by-step explanation:

2d - 6 = 5

+ 6 + 6

2d = 11

/2 /2

d = 11/2 or 5 1/2

7 0

3 years ago

Read 2 more answers