All categories

3 years ago

Pleas help I will give brainliest also please show how you did it please

Mathematics

2 answers:

lukranit [14]3 years ago

5 0

Answer:

n ≥ -1 or n ≤ 3

Combined:

-1 ≤ n ≤ 3

Since the inequalities are ≤, it'll be a closed circle

The closed circle will start AT -1 and another closed circle will end at 3

Serggg [28]3 years ago

4 0

Answer:

for n ≤ 3

we are going to put a circle on the 3 and make it filled in then you are going to start an arrow starting from the 3 going to the left all the way to the -10.

You might be interested in

Sign numbers, ratios & Proportion <br><br> (+10) + (+6) = ?

Viefleur [7K]

Answer:

Step-by-step explanation:

Putting a + in front of a number indicates that the number is positive. so we are just going to add it up.

10+6=16

10:6=5:3

Hope it helps:)

7 0

3 years ago

Read 2 more answers

A company uses three different assembly lines- A1, A2, and A3- to manufacture a particular component. Of thosemanufactured by li

hammer [34]

Answer:

The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

Step-by-step explanation:

The three different assembly lines are: A₁, A₂ and A₃.

Denote <em>R</em> as the event that a component needs rework.

It is given that:

$P (R|A_{1})=0.05\\P (R|A_{2})=0.08\\P (R|A_{3})=0.10\\P (A_{1})=0.50\\P (A_{2})=0.30\\P (A_{3})=0.20$

Compute the probability that a randomly selected component needs rework as follows:

$P(R)=P(R|A_{1})P(A_{1})+P(R|A_{2})P(A_{2})+P(R|A_{3})P(A_{3})\\=(0.05\times0.50)+(0.08\times0.30)+(0.10\times0.20)\\=0.069$

Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:

$P (A_{1}|R)=\frac{P(R|A_{1})P(A_{1})}{P(R)}=\frac{0.05\times0.50}{0.069} =0.3623$

Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

6 0

3 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

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:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago

A white tailed deer can sprint up to 30 miles per hour. American bison can run at speeds up to 3,520 feet per minute. Which anim

mario62 [17]

Answer:

The American bison is 10 miles per hour faster. It can run at 40 miles per hour.

Step-by-step explanation:

See the picture for the conversion. I used unit multiplies to change feet to miles and seconds to minutes.

4 0

1 year ago

The average weight of the top 5 fish caught at a fishing tournament was 12.3 pounds. Some of the weights of the fish are shown i

UNO [17]

Answer:

14.6

Step-by-step explanation:

It is given that,

The average weight of the top 5 fish caught at a fishing tournament was 12.3 pounds. From the attached figure, the weight of 5 fish are given. We need to find the weight of Wayne S. fish.

Average = sum of terms/no of terms

Let the weight of Wayne S. is x. So,

Here the sum of terms is x+12.8+12.6+11.8+9.7 and the number of terms is 5.

$12.3=\dfrac{x+12.8+12.6+11.8+9.7}{5}\\\\61.5=x+12.8+12.6+11.8+9.7\\\\61.5=x+46.9\\\\x=14.6$

So, the weight of the heaviest fish is 14.6.

3 0

3 years ago