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4 years ago

Is it possible to construct a triangle with side lengths of 6 cm, 11 cm, and 13 cm?

2 answers:

8 0

$a=6 \ cm, \ \ b=11 \ cm, \ \ c= 13 \cm\\\\Formula. \\It \ is \ a \ eries \ of \ three \ inequalities:\\ a+b>c, \\a+c>b, \\ b+c>a.$

$6+11 >13, \\6+13> 11, \\ 11+13> 6 \\ \\ 17 >13, \\19> 11, \\ 24> 6\\\\Answer: \ it \ is \ possible \ to \ construct \ a \ triangle \ with \ given \ sides$

vazorg [7]4 years ago

3 0

Sure. Why not ?

-- The 6 and the 11 can cover the 13 with some left over.

-- The 6 and the 13 can cover the 11 with some left over.

-- The 11 and the 13 can cover the 6 with a lot left over.

You might be interested in

How many 6” x 4” x 2” bricks will you need to construct a 14’5” x 5’5” x 0.5’ wall?

IRISSAK [1]

Answer:

1406 bricks

Step-by-step explanation:

We are given that the dimension of the bricks are,

Length = 6 inches, Width = 4 inches and Height = 2 inches.

Thus, we have,

Volume of 1 brick, $V_{B}$ = Length × Width × Height

i.e. Volume of 1 brick, $V_{B}$ = 6 × 4 × 2 = 48 inch²

So, the volume of 1 brick is 48 inch².

Also, the dimensions of the wall are given by,

Length = 14’5” = 173 inches, Width = 5’5” = 65 inches and Height = 0.5' = 6 inches

Volume of the wall, $V_{W}$ = Length × Width × Height

i.e. Volume of the wall, $V_{W}$ = 173 × 65 × 6 = 67470 inch²

So, the volume of the wall is 67470 inch².

Thus, we get,

Number of bricks = $\frac{V_{W}}{V_{B}}$

i.e. Number of bricks = $\frac{67470}{48}$

i.e. Number of bricks = 1406

Hence, 1406 bricks are needed to construct the wall.

5 0

3 years ago

The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development

Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

b) $X \sim Bin(n =20, p=0.8)$

c) $P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

d) $P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

e) $E(X) = np = 20*0.8 = 16$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

5 0

3 years ago

Please help! Correct answer only!

Irina-Kira [14]

Answer:

<em>" Expected Payoff " ⇒ $ 1.80 ; Type in 1.80</em>

Step-by-step explanation:

Take the probability of winning into consideration;

$Total Number of Tickets - 500,\\Tickets 1 Person Can Enter - 1 Ticket,\\\\Probability of Winning - 1 / 500,\\Money Won - 900 Dollars,\\\\Proportionality - 1 / 500 = x / 900 - where, x = " Expected Payoff "\\1 / 500 = x / 900 - CrossMultiplication,\\\\500 * x = 900,\\x = 900 / 500,\\x = 1.80 Dollars Won!\\\\Conclusion ; x = 1.80 Dollars$

<em>Solution ; " Expected Payoff " ⇒ $ 1.80</em>

3 0

3 years ago

A city pool has 3 wading pools. Each pool is in the shape of a parallelogram with a

Vikki [24]

Answer:

i think it's 13.44

Step-by-step explanation:

im not really sure, but let me know if im wright pls!!

hope this helped!!

6 0

3 years ago

Suzanne has one 10p coin one 50p coin and some 20p coins altogether she has £1.40 how many 20p coins does she have

mash [69]

Answer:

She has FOUR 20p coins altogether.

Step-by-step explanation:

1.40-0.10-0.50=0.8

0.8/0.2=4

6 0

3 years ago