All categories

3 years ago

Can you construct a triangle that has side lengths 8 m 10 m and 19 m

Mathematics

1 answer:

Alexxandr [17]3 years ago

4 0

Answer:

Step-by-step explanation:

Draw a triangle and label the sides a, b and c. Now we must know the rule you can use for drawing any triangle:

A+B> c

A+C>B

B+C>A

Sub you numbers into the rule:

8+10, is it greater that 19?

No, 8+10=18 and 18<19

This is proof that you can not draw a triangle with the side lengths of 8m, 10m and 19m.

You might be interested in

Factor n^3 - n^2 + 3n - 3.

lozanna [386]

N^2(n - 1) + 3(n - 1)

(n^2 + 3)(n - 1)

the answer is c

5 0

2 years ago

PLEASE HELP! ! ! urgent!

sertanlavr [38]

Answer:

$6(\frac{1}{3} \times\frac{7}{5} )=(6\times\frac{1}{3} )\frac{7}{5}$

$Answer: B$

Hope it helps...

Have a great day!!

8 0

3 years ago

Naval intelligence reports that 99 enemy vessels in a fleet of 1818 are carrying nuclear weapons. If 99 vessels are randomly tar

oee [108]

Answer:

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

Step-by-step explanation:

The vessels are destroyed and then not replaced, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

In this question:

Fleet of 18 means that $N = 18$

9 are carrying nuclear weapons, which means that $k = 9$

9 are destroyed, which means that $n = 9$

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,18,9,9) = \frac{C_{9,0}*C_{9,9}}{C_{18,9}} = 0.000021$

$P(X = 1) = h(1,18,9,9) = \frac{C_{9,1}*C_{9,8}}{C_{18,9}} = 0.001666$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.000021 + 0.001666 = 0.001687$

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

4 0

3 years ago

Mark the sum as either positive or negative

tester [92]

Answer:

1: Positive

2: Negative

3: Negative

4: Negative

Step-by-step explanation:

Positive plus Negative equals negative.

Positive plus positive or negative plus negative equals positive.

Hope this helps.

6 0

2 years ago

Read 2 more answers

14/2 three equivalent ratios

Anna71 [15]

Answer:

Three equivalent ratios of $\frac{14}{2}$ are $\frac{7}{1}$ , $\frac{21}{3}$ and $\frac{28}{8}$ .

Step-by-step explanation:

We are given a fraction $\frac{14}{2}$ .

We need to find the three equivalent ratios/fractions of $\frac{14}{2}$ .

The common factor of 14 and 2 is 2.

So, dividing top and bottom by 2, we get

14÷2 = 7 and 2÷2 that is $\frac{7}{1}$ .

Multiplying $\frac{7}{1}$ fraction by a common number 3 in top and bottom, we get

7 × 3 = 21

1 × 3 = 6

So, we get another equivalent fraction $\frac{21}{6}$ .

Multiplying $\frac{7}{1}$ fraction by a common number 3 in top and bottom, we get

7 × 4 = 28

1 × 4 = 4

So, we get another equivalent fraction $\frac{28}{4}$ .

Therefore, three equivalent ratios of $\frac{14}{2}$ are $\frac{7}{1}$ , $\frac{21}{3}$ and $\frac{28}{4}$ .

6 0

3 years ago