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2 years ago

8

The radius of a circular clock is 10.5 inches. Which measurement is the closet to the measurement of the circumference of the cl

ock in inches?

1 answer:

lys-0071 [83]2 years ago

8 0

Answer:

66 in approx

Step-by-step explanation:

Given data

Radius= 10.5 inches

The circumference is given as

C= 2πr

Substitute

C=2*3.142*10.5

C=65.982

Hence the Circumference is 66 in approx

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Team 2, serves the ball at 2-0-2, Team 1 returns the ball, and Team 2 volleys the ball with no return past Team 1. What is the c

Oksana_A [137]

Answer:

23

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8 0

2 years ago

9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operat

Marysya12 [62]

Answer:

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 2000, p = 1-0.995 = 0.005$

Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So

$P(X < 5) + P(X \geq 5) = 1$

We want $P(X \geq 5)$

So

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044$

$P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445$

$P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235$

$P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480$

$P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765$

$P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710$

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

4 0

3 years ago

RIDDLE ME THIS:

Finger [1]

Second place, because you are over taking the person that was in second so you are practically stealing the position from him

7 0

2 years ago

In ΔTUV, t = 410 cm, ∠U=27° and ∠V=78°. Find the length of v, to the nearest 10th of a centimeter.

telo118 [61]

Answer:

$v \approx 415.2\,cm$

Step-by-step explanation:

The angle T is:

$\angle T = 180^{\circ} - 27^{\circ} - 78^{\circ}$

$\angle T = 75^{\circ}$

Now, the length of v is determine by the Law of Sines:

$\frac{410\,cm}{\sin 75^{\circ}} = \frac{v}{\sin 78^{\circ}}$

$v = (410\,cm)\cdot \left(\frac{\sin 78^{\circ}}{\sin 75^{\circ}} \right)$

$v \approx 415.2\,cm$

6 0

3 years ago

Estimate 1 13 cos(x2) dx 0 using the Trapezoidal Rule and the Midpoint Rule, each with n = 4. (Round your answers to six decimal

babunello [35]

Answer:

(a) 4.152698

(b) 3.215557

Step-by-step explanation:

(a)

$\int\limits^{13}_1 {cos(x^2)} \, dx =M_n=$\sum_{n=1}^{\infty} f(m_i)\Delta x $$

n=4, so :

Each subinterval has length :

$\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3$

Therefore the subintervals consist of:

$[1,5], [5,9], [9,13]$

Now, the midpoints of these subintervals are:

$\frac{1+5}{2} =3\\\\\frac{5+9}{2} =7\\\\\frac{9+13}{2} =11$

Hence:

$M_4= 3*(cos(3^2))+3*(cos(7^2))+3*cos((11^2))\approx 4.152698$

(b)

$\int\limits^{13}_1 {cos(x^2)} \, dx =T_n=\frac{\Delta x}{2} (f(x_o)+2f(x_1)+2f(x_2)+...+2f(x_n_-_1)+f(x_n))$

n=4, so :

Each subinterval has length :

$\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3$

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

The endpoints of the subintervals consist of:

5,9

Hence:

$T_4= \frac{3}{2}(cos(1^2)+2*cos(5^2)+2*cos(9^2)+cos(13^2)) \approx 3.215557$

8 0

3 years ago