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2 years ago

PLSSSS BELP IF YOU TO TURLY KNOW THISSS

Mathematics

2 answers:

GalinKa [24]2 years ago

5 0

Answer:

Estimate

mark me as brainliest plz

VLD [36.1K]2 years ago

5 0

Answer:

Exact:

she would give her back 6.93

Estimate:

7.00

Step-by-step explanation:

Exact!

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What are the coordinates of the endpoints of the midsegment for △LMN that is parallel to LN¯¯¯¯¯ ?

Brut [27]

(3, 4.5) and (3, 3)

The midsegment of a triangle is a line connecting the midpoints of two sides of the triangle. So a triangle has 3 midsegments. Since you want the midsegment that's parallel to LN, we need to select the midpoints of LM and MN. The midpoint of a line segment is simply the average of the coordinates of each end point of the line segment. So:

Midpoint LM:

((0+6)/2, (5+4)/2) = (6/2, 9/2) = (3, 4.5)

Midpoint MN:

((6+0)/2, (4+2)/2) = (6/2, 6/2) = (3, 3)

So the desired end points are (3, 4.5) and (3, 3)

3 0

3 years ago

Order these numbers from least to greatest. 11 , 3.30 , 175 , −10

timurjin [86]

If you would like to order the following numbers from least to greatest, you can do this using the following steps:

11, 3.30, 175, - 10
- 10 < 3.30 < 11 < 175

The correct result would be - 10 < 3.30 < 11 < 175.

5 0

3 years ago

Industry standards suggest that 8 percent of new vehicles require warranty service within the first year. Jones Nissan in Sumter

jenyasd209 [6]

Answer:

Step-by-step explanation:

Let X be the no of vehicles that require warranty service within I year out of vehicles sold (9) yesterday.

Each vehicle is independent of the other to get warranty service

Hence X is binomial with p=8% = 0.08

So P(X=x) = $9Cx (0.08)^x (1-0.08)^{9-x} ,x=0,1,2...9$

$a) P(X=0) = 0.472161\\b) P(x=1) = 0.3695$

c) $P(X=2) = 0.1285$

d) Mean = Mean of binomial = np =

$9(0.08) = 0.72$

Var (x) = npq = $0.6624$

Std dev = square root of variance = 0.8139

5 0

3 years ago

Read 2 more answers

What is a triangle with at least two congruent sides called?

PilotLPTM [1.2K]

Isosceles triangle I think

3 0

2 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

3 years ago