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8090 [49]
2 years ago
9

You’re given two side lengths of 3 centimeters and 5 centimeters. Which measurement can you use for the length of the third side

to construct a valid triangle?
Mathematics
1 answer:
mrs_skeptik [129]2 years ago
8 0

Answer:

You could use a measurement of 4 centimeters.

The valid range is  2  < x < 8  cms.

Step-by-step explanation:

The third side must either be less than 3+5, that is less than 8 cms and  it must be greater than (5-3) = 2 cms. If it is any other length we could not construct a triangle.

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You decide to make milk a loss leader. You drop the price from its cost of $2.59 per gallon to $1.99 per gallon. What is the per
Veseljchak [2.6K]

Answer:

<u>The percentage loss per gallon is 23.17%</u>

Step-by-step explanation:

Original price per gallon = $ 2.59

Discounted price per gallon = $ 1.99

Now, let's calculate the percentage loss per gallon, using the Direct Rule of Three, as follows:

Price      Percentage

2.59           100

1.99               x

*****************************

2.59 * x = 1.99 * 100

2.59x = 199

x = 199/2.59

x = 76.83

Percentage loss = 100 - 76.83

<u>Percentage loss = 23.17%</u>

4 0
2 years ago
Read 2 more answers
Let’s assume the following statements are true: Historically, 75% of the five-star football recruits in the nation go to univers
marishachu [46]
<span>Given:
75% of the five-star football recruits in the nation go to universities in the three most competitive athletic conferences. </span>→ 25% goes to other schools.
<span>
five-star recruits get full football scholarships 93% of the time, regardless of which conference they go to. </span>→ 7% of the 5-star recruits don't get full football scholarships.<span>

a. The probability that a randomly selected five-star recruit who chooses one of the best three conferences will be offered a full football scholarship? 
75% * 93% = 69.75%

b. What are the odds a randomly selected five-star recruit will not select a university from one of the three best conferences?
25% of selected five-star recruit will not select a university from one of the three best conferences. I got the number based on the given data. Since, 75% will go, the remaining percent won't go. Total percentage should be 100% of the population. 

c. Explain whether these are independent or dependent events. Are they Inclusive or exclusive? 
These are independent events. One can still go to different school and still be legible for the full football scholarship. 

For question 2, pls. see attachment.</span>

4 0
2 years ago
Determine whether the lines are parallel, intersect, or coincide. x-5y=0, y+1=1/5(x+5)
MAXImum [283]

<u>Answer-</u>

<em>The lines are two </em><em>coinciding lines</em><em> or the </em><em>same lines</em><em>.</em>

<u>Solution-</u>

The given line equations are

the first one,

\Rightarrow x-5y=0  

the second one,

\Rightarrow y+1=\dfrac{1}{5}(x+5)\\\\\Rightarrow 5y+5=x+5\\\\\Rightarrow 5y=x\\\\\Rightarrow x-5y=0

As we know two line equations A_1x+B_1y+C_1=0 and A_2x+B_2y+C_2=0 will be,

  1. Parallel if, \dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}
  2. Coincide if, \dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}=\dfrac{C_1}{C_2}
  3. Intersect if, \dfrac{A_1}{A_2}\neq \dfrac{B_1}{B_2}

As here,

\Rightarrow \dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}=\dfrac{C_1}{C_2}

\Rightarrow \dfrac{1}{1}=\dfrac{-5}{-5}

(C₁ and C₂ aren't considered as they are 0)

Therefore, the lines are two coinciding lines or the same lines.

4 0
2 years ago
Suppose that the travel time from your home to your office is a normal random variable with mean 40 minutes and standard deviati
Degger [83]

Answer:

See explaination

Step-by-step explanation:

Given in the question

travel mean time from your home on monday to thursday = 40 minutes

Standard deviation = 7 minutes

travel mean time from your home on friday = 35 minutes

Standard deviation = 5 minutes

(a)

Office meeting time on Monday = 1PM

He want to 95 percent certain that you will not be late for an office meeting at 1PM on monday

So p-value = 0.95 and Z-score from Z table is 1.645

So the latest time at which you should leave home can be calculated as

X = Mean + Z-score *Standard deviation = 40 + 1.645*7 = 40+11.515 = 51.515 or 52 minutes before the time

So latest time at which you should leave home = 52 minutes before 1 PM i.e. 12:08 PM

(b)

Shift time = 8AM

Leaving time = 7:10 AM

So Leaving before time = 50 minutes

Total working days = 250

So Total Monday to thursday = 200

Total friday = 50

Probability of getting late on Monday to Thursday can be calculated using standard normal curve as follows:

Z-score = (X - Mean)/SD = (50-40)/7 = 1.43

From Z table, we found probability of getting late = 0.0764

So Total number of late days in Monday to Thursday = 200*0.0764 = 15.3 or 16 days

Probability of getting late on Friday can be calculated using standard normal curve as follows:

Z-score = (X - Mean)/SD = (50-35)/5 = 3

From Z table, we found probability of getting late on friday= 0.00135

So Total number of late days in Friday = 50*0.00135 = 0.0675 or 1 day

So total late days = 16 +1 =17 days

So Its correct answer is C. i.e. 12.08 and 17

8 0
2 years ago
How many numbers are 10 units from 0 on the number line? Type your answer as a numeral.How many numbers are 10 units from 0 on t
hammer [34]
2 numbers are 10 units from 0.  If you start from zero and count 10 units  to the right, you should be at a positive number.  But, you can also move 10 units to the left. If you do this, you should be at a negative number that  is ALSO 10 units away from 0, just in a different direction than positive number. both are 10 units from 0.   There always 2 numbers, one 10 units to the right of zero, and one 10 units to the left of zero. Hope this helped!  Tell me if I'm wrong or right. :)
                
7 0
2 years ago
Read 2 more answers
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