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3 years ago

Please help. I don't understand

Mathematics

1 answer:

Andreas93 [3]3 years ago

5 0

Answer:

Length of third side is: $\mathbf{8m^3-4m^2+3m}$

Step-by-step explanation:

Perimeter of triangle = $18m^3+2m^2-m$

Length of side 1= $5m^3+3m^2-2m$

Length of side 2= $5m^3+3m^2-2m$

We need to find length of third side.

The formula used is: $Perimeter \:of\:triangle=Sum\:of\:all\:sides$

Putting values and finding length of third side.

$Perimeter \:of\:triangle=Sum\:of\:all\:sides\\18m^3+2m^2-m=5m^3+3m^2-2m+5m^3+3m^2-2m+Length\:of\:3rd\:side\\18m^3+2m^2-m=5m^3+5m^3+3m^2+3m^2-2m-2m+Length\:of\:3rd\:side\\18m^3+2m^2-m=10m^3+6m^2-4m+Length\:of\:3rd\:side\\Length\:of\:3rd\:side=18m^3+2m^2-m-(10m^3+6m^2-4m)\\Length\:of\:3rd\:side=18m^3+2m^2-m-10m^3-6m^2+4m\\Length\:of\:3rd\:side=18m^3-10m^3+2m^2-6m^2-m+4m\\Length\:of\:3rd\:side=8m^3-4m^2+3m$

So, Length of third side is: $\mathbf{8m^3-4m^2+3m}$

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In an arithmetic sequence, the nth term an is given by the formula an=a1+(n−1)d, where a1 is the first term and d is the commo

PilotLPTM [1.2K]

Answer:

105th term of given series is

$a_n=\dfrac{105}{2}$

Step-by-step explanation:

Given series is

$\dfrac{1}{2},\ 1,\ \dfrac{3}{2},\ 2,\ \dfrac{5}{2}.....$

As we can see,

$\textrm{First term},a_1=\dfrac{1}{2}$

Also,

$1-\dfrac{1}{2}=\dfrac{3}{2}-1=2-\dfrac{3}{3}=.....=\dfrac{1}{2}$

hence, we can say given series is in arithmetic progression,

with common difference,

$d=\ \dfrac{1}{2}$

As given in question the nth term in A.P is given by

$a_n=a_1+(n-1)d$

since we have to find the 105th term, so we can write

$a_{105}=\dfrac{1}{2}+(105-1)\dfrac{1}{2}$

$=\dfrac{1}{2}+\dfrac{104}{2}$

$=\dfrac{105}{2}$

Hence, the 105th term of given series of A.P is $\dfrac{105}{2}$ .

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3 years ago

How do we find the quadratic equation in standard form with the given roots of -1/3, 5?

Olin [163]

I think this is the answer. Because it doesn’t mention a leading coefficient but it’s a quadratic function so it would be
1(x-1/3)(x-5) and that should be your answer. The simplified version is down below

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3 years ago

Find the area of the shaded regions: the green is the shaded area

algol13

Answer:

the green area stands for the safe operation space

Step-by-step explanation:

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2 years ago

A bag contains 4 blue marbles and 2 yellow marbles. Two marbles are randomly chosen (the first marble is NOT replaced before dra

VMariaS [17]

Answer:

0.4 = 40% probability that both marbles are blue.

0.0667 = 6.67% probability that both marbles are yellow.

53.33% probability of one blue and then one yellow

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

Hypergeometric distribution:

The probability of x sucesses 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 sucesses.

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.

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

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

What is the probability that both marbles are blue?

4 + 2 = 6 total marbles, which means that $N = 6$

4 blue, which means that $k = 4$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

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

$P(X = 2) = h(2,6,2,4) = \frac{C_{4,2}*C_{2,0}}{C_{6,2}} = 0.4$

0.4 = 40% probability that both marbles are blue

What is the probability that both marbles are yellow?

4 + 2 = 6 total marbles, which means that $N = 6$

2 yellow, which means that $k = 2$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

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

$P(X = 2) = h(2,6,2,2) = \frac{C_{2,2}*C_{4,0}}{C_{6,2}} = 0.0667$

0.0667 = 6.67% probability that both marbles are yellow.

What is the probability of one blue and then one yellow?

Total is 100%.

Can be:

Both blue(40%)

Both yellow(6.67%)

One blue and one yellow(x%). So

40 + 6.67 + x = 100

x = 100 - 46.67

x = 53.33

53.33% probability of one blue and then one yellow.

If you are told that both selected marbles are the same color, what is the probability that both are blue?

Conditional probability.

Event A: Both same color

Event B: Both blue

Probability of both being same color:

Both blue(40%)

Both yellow(6.67%)

This means that $P(A) = 0.4 + 0.0667 = 0.4667$

Probability of both being the same color and blue:

40% probability that both are blue, which means that $P(A \cap B) = 0.4$

Desired probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.4667} = 0.8571$

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

8 0

2 years ago

9^3/9^9<br> What does this equation equal?

ololo11 [35]

I'm assuming the problem is nine to the third power divided by nine to the ninth power, so the answer would be 0.000001882. Fraction form would be 1/531441.

Sorry about the first answer, I read the problem wrong.

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2 years ago

Read 2 more answers

Please help. I don't understand​

Please help. I don't understand