Prove that a triangle with two congruent altitudes is an isoceles triangle.

What is 30% of 70? O 10 O 14 O 18 O 21 O 28

Sedaia [141]

Answer:

your answer is 21

Step-by-step explanation:

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7 0

2 years ago

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In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz

Ahat [919]

Answer:

a) There is a 18.75% probability that the first question that she gets right is the second question.

b) There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

c) There is a 10.35% probability that she gets the majority of the questions right.

Step-by-step explanation:

Each question can have two outcomes. Either it is right, or it is wrong. So, for b) and c), we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

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

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

And $\pi$ is the probability of X happening.

In this problem we have that:

Each question has 4 choices. So for each question, Robin has a $\frac{1}{4} = 0.25$ probability of getting ir right. So $\pi = 0.25$ . There are five questions, so $n = 5$ .

(a) What is the probability that the first question she gets right is the second question?

There is a 75% probability of getting the first question wrong and there is a 25% probability of getting the second question right. These probabilities are independent.

So

$P = 0.75(0.25) = 0.1875$

There is a 18.75% probability that the first question that she gets right is the second question.

(b) What is the probability that she gets exactly 1 or exactly 2 questions right?

This is: $P = P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955$

$P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637$

$P = P(X = 1) + P(X = 2) = 0.3955 + 0.2637 = 0.6592$

There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

(c) What is the probability that she gets the majority of the questions right?

That is the probability that she gets 3, 4 or 5 questions right.

$P = P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 3) = C_{5,3}.(0.25)^{3}.(0.75)^{2} = 0.0879$

$P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146$

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

$P = P(X = 3) + P(X = 4) + P(X = 5) = 0.0879 + 0.0146 + 0.001 = 0.1035$

There is a 10.35% probability that she gets the majority of the questions right.

6 0

3 years ago

*Silly and/or spam answers will not be tolerated*

stepladder [879]

Answer:

-1/8

Step-by-step explanation:

lim x approaches -6 (sqrt( 10-x) -4) / (x+6)

Rationalize

(sqrt( 10-x) -4) (sqrt( 10-x) +4)

------------------- * -------------------

(x+6) (sqrt( 10-x) +4)

We know ( a-b) (a+b) = a^2 -b^2

a= ( sqrt(10-x) b = 4

(10-x) -16

-------------------

(x+6) (sqrt( 10-x) +4)

-6-x

-------------------

(x+6) (sqrt( 10-x) +4)

Factor out -1 from the numerator

-1( x+6)

-------------------

(x+6) (sqrt( 10-x) +4)

Cancel x+6 from the numerator and denominator

-1

-------------------

(sqrt( 10-x) +4)

Now take the limit

lim x approaches -6 -1/ (sqrt( 10-x) +4)

-1/ (sqrt( 10- -6) +4)

-1/ (sqrt(16) +4)

-1 /( 4+4)

-1/8

5 0

3 years ago

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Verify that the given value is the solustion of the equation x=6 for 2z-4=8

klemol [59]

Answer:

8-4=2z.. 2z+4=6 so x=6

6 0

3 years ago

How many solutions are there to the system of equations? StartLayout enlarged left-brace 1st row 4 x minus 5 y = 5 2nd row negat

kkurt [141]

Answer:

Only one solution $x=\frac{5}{4},y=0$

Step-by-step explanation:

$4x-5y=5..............(1)\\0.08x+0.10y=0.10...........(2)$

Multiply equation $(2)$ by $50$

$4x-5y=5........(3)$

Now from equation $(1)$ $+$ equation $(3)$

$8x=10\\x=\frac{10}{8}\\x=\frac{5}{4}$

Put the value in equation $(1)$

$5-5y=5\\5y=0\\y=0$

hence only one solution exists $\left(\frac{5}{4},0\right)$ .

3 0

3 years ago

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