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2 years ago

If m∠EBD= 4x - 8 and m∠EBC= 5x + 20, find the value of x and m∠EBC.

Mathematics

1 answer:

coldgirl [10]2 years ago

4 0

Answer:

-28 = x

Step-by-step explanation

[5x + 20] + [4x - 8] = 180

9x + 12 = 180

- 12 -12

______________

9x = 168

18⅔ = x

m∠EBC = 5[18⅔] + 20 = 113⅓

93⅓

You will get a negative answer if you set them equal to each other, which is not allowed when you are talking about length.

I hope this is correct, and as always, I am joyous to assist anyone at any time.

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Answer:

The probability that exactly eight of them take more than 93.6 minutes is 5.6015 $\times 10^{-6}$ .

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We are given that it is known that times for service calls follow a normal distribution with a mean of 75 minutes and a standard deviation of 15 minutes.

A random sample of twelve service calls is taken.

So, firstly we will find the probability that service calls take more than 93.6 minutes.

Let X = times for service calls.

So, X ~ Normal( $\mu=75,\sigma^{2} =15^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean time = 75 minutes

$\sigma$ = standard deviation = 15 minutes

Now, the probability that service calls take more than 93.6 minutes is given by = P(X > 93.6 minutes)

P(X > 93.6 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{93.6-75}{15}$ ) = P(Z > 1.24) = 1 - P(Z $\leq$ 1.24)

= 1 - 0.8925 = 0.1075

The above probability is calculated by looking at the value of x = 1.24 in the z table which has an area of 0.8925.

Now, we will use the binomial distribution to find the probability that exactly eight of them take more than 93.6 minutes, that is;

$P(Y = y) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; y = 0,1,2,3,.........$

where, n = number of trials (samples) taken = 12 service calls

r = number of success = exactly 8

p = probability of success which in our question is probability that

it takes more than 93.6 minutes, i.e. p = 0.1075.

Let Y = Number of service calls which takes more than 93.6 minutes

So, Y ~ Binom(n = 12, p = 0.1075)

Now, the probability that exactly eight of them take more than 93.6 minutes is given by = P(Y = 8)

P(Y = 8) = $\binom{12}{8}\times 0.1075^{8} \times (1-0.1075)^{12-8}$

= $495 \times 0.1075^{8} \times 0.8925^{4}$

= 5.6015 $\times 10^{-6}$ .

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