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skelet666 [1.2K]
3 years ago
13

What is the area of the triangle?

Mathematics
2 answers:
Alika [10]3 years ago
7 0

Answer:

15 I think

Step-by-step explanation:

Strike441 [17]3 years ago
6 0

Answer:

15

Step-by-step explanation:

if right pls mark brainlest

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Explain the systematic trial method for solving equations
adoni [48]
Systematic sampling is a type of probability sampling method in which sample members from a larger population are selected according to a random starting point but with a fixed, periodic interval. This interval, called the sampling interval, is calculated by dividing the population size by the desired sample size.
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1.62 as a percentage
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1.62 as a percentage would be 162%. Hope this is helpful!
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One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
An investigator wants to assess whether the mean m = the weight of passengers flying on small planes exceeds the FAA guideline o
stira [4]

Answer:

H_{0}: \mu= 185 and H_{a}: \mu > 185

Step-by-step explanation:

The null hypothesis H_{0} states that a population parameter (such as the mean, the standard deviation, and so on) is equal to a hypothesized value. We can write the null hypothesis in the form H_{0}: parameter = value

In this context, the investigator's null hypothesis should be that the average total weight is no different than the reported value by the FAA. We can write it in this form H_{0}: \mu= 185.

The alternative hypothesis H_{a} states that a population parameter is smaller, greater, or different than the hypothesized value in the null hypothesis. We can write the alternative hypothesis in one of three forms

H_{a}: parameter > value\\H_{a}: parameter < value\\H_{a}: parameter \neq value

The investigator wants to know if the average weight of passengers flying on small planes exceeds the FAA guideline of the average total weight of 185 pounds. He should use H_{a}: \mu > 185 as his alternative hypothesis.

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3 years ago
Claire is going to invest $2,600 and leave it in an account for 18 years. Assuming the interest is compounded continuously, what
Bogdan [553]

Answer:

Rate of interest r = 2.83 % (Approx.)

Step-by-step explanation:

Given:

Amount invested p = $2,600

Amount get A = $4,300

Number of year n = 18

Find:

Rate of interest r

Computation:

A = p(1+r)ⁿ

4,300 = 2,600(1+r)¹⁸

(1+r)¹⁸ = 1.653846

Rate of interest r = 2.83 % (Approx.)

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3 years ago
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