Ask question

Ask question

All categories

1 year ago

7

In triangle ABC, m∠BAC = 50°. If m∠ACB = 30°, then the triangle is triangle. If m∠ABC = 40°, then the triangle is triangle. If t

riangle ABC is isosceles, and AB = 6 and BC = 4, then AC =

2 answers:

Agata [3.3K]1 year ago

8 0

Acb is an OBTUSE triangle.
Abc is a RIGHT triangle.
Then AC= 4 or 6
these are the right answers for edquity

frosja888 [35]1 year ago

5 0

Answer:

OBTUSE, RIGHT, and 4 or 6

Step-by-step explanation:

You might be interested in

Which describes money that is received from sales of goods or services?

Lana71 [14]

the answer would be revenue

7 0

2 years ago

Cindy saves $420 each month . her monthly salary is $ 2100. what is the ratio of the amount cindy saves to the amount she earns

AfilCa [17]

I took the test the correct answer is 1:5.

7 0

1 year ago

Which polynomial function could be represented by the graph below?

Svetllana [295]

Note that

1.

$f(x)=x^3+x^2-6x=x(x^2+x-6)=x(x-2)(x+3)$

The x-intercepts are at points x=-3, x=0, x=2. The graph should be increasing - decreasing - increasing.

2.

$f(x)=x^3-x^2-6x=x(x^2-x-6)=x(x+2)(x-3)$

The x-intercepts are at points x=-2, x=0, x=3. The graph should be increasing - decreasing - increasing.

3.

$f(x)=-2x^3-2x^2+12x=-2x(x^2+x-6)=-2x(x-2)(x+3)$

The x-intercepts are at points x=-3, x=0, x=2. The graph should be decreasing - increasing - decreasing.

4.

$f(x)=-2x^3+x^2+12x=-2x(x^2-x-6)=-2x(x+2)(x-3)$

The x-intercepts are at points x=-2, x=0, x=3. The graph should be decreasing - increasing - decreasing.

From the diagram you can see that x-intercepts are at points x=-3, x=0, x=2 and the graph is decreasing-increasing-decreasing.

Answer: correct choice is 3.

7 0

2 years ago

Read 2 more answers

A house is purchased for $300,000 and its value appreciates by 2.75% per year.In how many years will the house be worth $375,000

MArishka [77]

Answer:

In 9.09 years

Step-by-step explanation:

4 0

1 year ago

Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

4 of the next 20 drivers do you expect to come to a complete stop

4 0

2 years ago