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IgorLugansk [536]
2 years ago
8

Question 5 30 points

Mathematics
1 answer:
IRINA_888 [86]2 years ago
5 0

Answer:

y = 1/2x + 2

Explanation:

The line intercepts the y-axis at 2, so that would be "b" using the slope-intercept formula y = mx + b. (I won't explain the slope as I am tired now.)

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-3p-4<6 help me please never done these before and teacher isn’t at school to ask!!!
tia_tia [17]

Answer:

p>-1/3

Step-by-step explanation:

-3p-4<6

+4 +4

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

-3p<10

Divide -3 on both sides and you get something like p>-1/3 because you have to flip the sign.

7 0
1 year ago
Help me prettyyyy please
OlgaM077 [116]

Answer:

Hii! Surface area= 207.35, volume= 226.19. Hope this helps!

3 0
3 years ago
Read 2 more answers
Uranus moves in an elliptical orbit with the sun at one of the foci. The length of the half of the major axis is 2,876,769,540 k
Alina [70]

Answer:

The minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

Step-by-step explanation:

Consider the provided information.

The length of the half of the major axis is 2,876,769,540 kilometers, and the eccentricity is 0.0444.

The eccentricity (e) of an ellipse is the ratio of the distance from the center to the foci (c) and the distance from the center to the vertices (a).

e=\frac{c}{a}

Substitute a = 2,876,769,540 and e = 0.0444 in above formula and solve for c.

0.0444=\frac{c}{2,876,769,540 }

c=127728567.576

Minimum distance of Uranus from the sun is:

a-c=2,876,769,540-127728567.576\\a-c=2749040972.424\approx2749040972

Hence, the minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

6 0
3 years ago
HELP:How many ways can four
Tju [1.3M]

Answer:

24 ways.

Step-by-step explanation:

In this case, you just need a factorial.

For the first seat, you have 4 students you can place.

For the second seat, you have 3 students.

For the third, you have 2.

For the fourth, you have 1.

So, you can arrange the students by doing 4 * 3 * 2 * 1 = 12 * 2 = 24 ways.

Hope this helps!

8 0
2 years ago
Read 2 more answers
According to an​ airline, flights on a certain route are on time 80​% of the time. Suppose 17 flights are randomly selected and
tensa zangetsu [6.8K]

Answer:

a) Check Explanation

b) Probability that 11 out of the 17 randomly selected flights are on time = P(X = 11) = 0.0680

c) Probability that fewer than 11 out of the 17 randomly selected flights are on time

= P(X < 11) = 0.0377

d) Probability that at least 11 out of the 17 randomly selected flights are on time

= P(X ≥ 11) = 0.9623

e) Probability that between 9 and 11 flights, inclusive, out of the randomly selected 17 are on time = P(9 ≤ X ≤ 11) = 0.1031

Step-by-step explanation:

a) How to know a binomial experiment

1) A binomial experiment is one in which the probability of success doesn't change with every run or number of trials. (Probability of each flight being on time is 80%)

2) It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure. (It's either the flights are on time or not).

3) The outcome of each trial/run of a binomial experiment is independent of one another.

All true for this experiment.

b) Probability that exactly 11 flights are on time.

Let X be the random variable that represents the number of flights that are on time out of the randomly selected 17.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 17 randomly selected flights

x = Number of successes required = number of flights required to be on time

p = probability of success = Probability of a flight being on time = 80% = 0.80

q = probability of failure = Probability of a flight NOT being on time = 1 - p = 1 - 0.80 = 0.20

P(X = 11) = ¹⁷C₁₁ (0.80)¹¹ (0.20)¹⁷⁻¹¹ = 0.06803777953 = 0.0680

c) Probability that fewer than 11 flights are on time

This is also computed using binomial formula

It is the probability that the number of flights on time are less than 11

P(X < 11) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0376634429 = 0.0377

d) Probability that at least 11 out of the 17 randomly selected flights are on time

This is the probability of the number of flights on time being 11 or more.

P(X ≥ 11) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17)

= 1 - P(X < 11)

= 1 - 0.0376634429

= 0.9623365571 = 0.9623

e) Probability that between 9 and 11 flights, inclusive, are on time = P(9 ≤ X ≤ 11)

This is the probability that exactly 9, 10 or 11 flights are on time.

P(9 ≤ X ≤ 11) = P(X = 9) + P(X = 10) + P(X = 11)

= 0.0083528524 + 0.02672912767 + 0.06803777953

= 0.1031197592 = 0.1031

Hope this Helps!!!

3 0
2 years ago
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