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anastassius [24]
3 years ago
11

A repair person is scheduled to be at your apartment between 9:00am to 12:00pm but you have to leave at 11:30 to help out a frie

nd. If the repair person is equally likely to be at your apartment any time during the appointment time, what is the probability that you will be there?
Mathematics
2 answers:
sergiy2304 [10]3 years ago
7 0
<h3>Answer:</h3>

5/6

<h3>Explanation:</h3>

There are 6 half-hours between 9 and 12. You will be there for 5 of them. The probability you will be there is 5/6.

mixer [17]3 years ago
4 0

A repair person is scheduled to be at your apartment between 9:00am to 12:00pm but  you have to leave at 11:30 to help out a friend. If the repair person is equally likely to be  at your apartment any time during the appointment time, what is the probability that you  will be there?

83.3%

_______________________________________________

*100% CORRECT ANSWERS

Question 1  

What is the approximate Probability of drawing a RED FACE card from a standard deck  of shuffled cards.

11.5%

Question 2  

An opaque bag contains 5 green marbles, 3 blue marbles, and 2 red marbles. If a marble is drawn at random what are the ODDS of drawing a red marble?  

1:4

Question 3  

What is the probability of flipping a penny and a nickel and having them both land on  heads?

25%

Question 4

A solider is parachute training. He jumps out of a plane high enough that the pilot determines the solider will land somewhere in an 240 foot radius land zone. The pilow forgot to mention to the solider that the parachute has very little maneuverablity (no  steering). and there are 30 trees in the landing zone. Each tree from the air practically outlines a circle with a radius of 8 feet. What is the geometric probability that the solider  will land in a tree?

0.03333

Question 5  

A repair person is scheduled to be at your apartment between 9:00am to 12:00pm but  you have to leave at 11:30 to help out a friend. If the repair person is equally likely to be  at your apartment any time during the appointment time, what is the probability that you  will be there?

83.3%

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Two corresponding sides of two similar triangles are 3cm and 5cm. The area of the first triangle is 12cm2. What is the area of t
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Answer:

33\dfrac{1}{3}\ cm^2

Step-by-step explanation:

If two corresponding sides of two similar triangles are 3cm and 5cm, then the scale factor is

k=\dfrac{a_1}{a_2}=\dfrac{3}{5}.

Two similar triangles have their area proportional with the scale factor of k^2.

Hence,

\dfrac{A_1}{A_2}=k^2,\\ \\\dfrac{12}{A_2}=\left(\dfrac{3}{5}\right)^2,\\ \\\dfrac{12}{A_2}=\dfrac{9}{25},\\ \\A_2=\dfrac{12\cdot 25}{9}=\dfrac{100}{3}=33\dfrac{1}{3}\ cm^2.

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Six less than three - fifths of x<br><br> Translate the expression
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twinlakes on the shelf of a covenience store lose their fresh tastines over time. we say that the taste quality is 1 when the tw
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Answer:

The graph is shown below.

The time to make the taste to half is <u>4.265 s.</u>

Step-by-step explanation:

Given:

Initial value of the taste is, Q_0=1

Therefore, the quality of taste over time 't' is given as:

Q(t)=Q_0(0.85)^t\\Q(t)=1(0.85)^t

Now, when the taste reduces to half, Q=0.5

Therefore,

0.5=1(0.85)^t

Taking natural log on both the sides, we get:

\ln(0.5)=\ln(0.85)^t\\\ln(0.5)=t\ln(0.85)\\t=\frac{ln(0.5)}{\ln(0.85)}=4.265\ s

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4 0
3 years ago
<img src="https://tex.z-dn.net/?f=4x%20%2B%203y%20%3D%200%20%5C%5C5y%20%2B%2053%20%3D%2011x%20%5C%5C%20" id="TexFormula1" title=
Anvisha [2.4K]

Answer:

x = 3, y = -4

Step-by-step explanation by substitution:

Solve the following system:

{4 x + 3 y = 0 | (equation 1)

5 y + 53 = 11 x | (equation 2)

Express the system in standard form:

{4 x + 3 y = 0 | (equation 1)

-(11 x) + 5 y = -53 | (equation 2)

Swap equation 1 with equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

4 x + 3 y = 0 | (equation 2)

Add 4/11 × (equation 1) to equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+(53 y)/11 = -212/11 | (equation 2)

Multiply equation 2 by 11/53:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+y = -4 | (equation 2)

Subtract 5 × (equation 2) from equation 1:

{-(11 x)+0 y = -33 | (equation 1)

0 x+y = -4 | (equation 2)

Divide equation 1 by -11:

{x+0 y = 3 | (equation 1)

0 x+y = -4 | (equation 2)

Collect results:

Answer: {x = 3 , y = -4

_____________________________________

Solve the following system:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for x:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Isolate terms with x to the left hand side.

Subtract 3 y from both sides:

{4 x = -3 y

5 y + 53 = 11 x

Hint: | Solve for x.

Divide both sides by 4:

{x = -(3 y)/4

5 y + 53 = 11 x

Hint: | Perform a substitution.

Substitute x = -(3 y)/4 into the second equation:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for y:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Isolate y to the left hand side.

Subtract 53 - (33 y)/4 from both sides:

{x = -(3 y)/4

(53 y)/4 = -53

Hint: | Solve for y.

Multiply both sides by 4/53:

{x = -(3 y)/4

y = -4

Hint: | Perform a back substitution.

Substitute y = -4 into the first equation:

Answer: {x = 3 , y = -4

7 0
3 years ago
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