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zavuch27 [327]
3 years ago
11

Complete the statements to find the measurements of Za and Zb.

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0
What do you want me to find
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Natasha2012 [34]
Well, there are 52 cards in a deck, with 4 suits. there is a Jack and queen per suit, so that is a total of 8 jacks and queens in the deck. that probability looks like 8:52 as a ratio, or 2:13 simplified. so, you are likely to draw a Jack or queen 8 out of 52 times, or 2 out of 13 times
5 0
3 years ago
1400000000000000000000km in scientific notation
viva [34]

1 400 000 000 000 000 000 000 km = 1.4 * 10^21 km

8 0
3 years ago
Please help! will give brainliest
earnstyle [38]

Answer:

Step-by-step explanation:

y-2=3(x-5)

y-2=3x-15

Slope intercept form is y=3x-13

y-12= -2(x-3)

y-12= -2x+6

Slope intercept form is y= -2x+18

6 0
3 years ago
Which of the following are one-dimensional and have infinite length?
e-lub [12.9K]

Answer:

Answer is line and ray.

Step-by-step explanation:

Ray -  A ray starts from a point and ends to the infinity.

Therefore, ray is one dimensional and has infinite length.

Point - It is one dimensional but has no length.

Plane - It may be three/two dimensional.

Segment - It is one dimensional but has a limited length.

Line - It's one  dimensional with infinite length.

Angle - It's two dimensional structure.

Answer is line and ray.

5 0
3 years ago
2. Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,00
svlad2 [7]

Using the binomial distribution, it is found that:

a) There is a 0.0501 = 5.01% probability that you need to contact four people.

b) You expect to contact 1.82 students until you find one who lives within five miles of you.

c) The standard deviation is of 1.22 students.

d) There is a 0.3369 = 33.69% probability that 3 of them live within five miles of you.

e) It is expected that 2.75 students live within five miles of you.

For each student, there are only two possible outcomes. Either they live within 5 miles of you, or they do not. The probability of a student living within 5 miles of you is independent of any other student, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 55% of the students live within five miles of you, thus p = 0.55.

Item a:

This probability is P(X = 0) when n = 3(none of the first three living within five miles of you) multiplied by 0.55(the fourth does live within five miles), hence:

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

P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125

p = 0.091125(0.55) = 0.0501

0.0501 = 5.01% probability that you need to contact four people.

Item b:

The expected number of trials in the binomial distribution until q successes is given by:

E = \frac{q}{p}

In this problem, p = 0.55, and 1 trial, thus q = 1, hence:

E = \frac{1}{0.55} = 1.82

You expect to contact 1.82 students until you find one who lives within five miles of you.

Item c:

The standard deviation of the number of trials until q successes are found is given by:

S = \frac{\sqrt{q(1 - p)}}{p}

Hence, since q = 1, p = 0.55:

S = \frac{\sqrt{0.45}}{0.55} = 1.22

The standard deviation is of 1.22 students.

Item d:

This probability is P(X = 3) when n = 5, hence:

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

P(X = 3) = C_{5,3}.(0.55)^{3}.(0.45)^{2} = 0.3369

There is a 0.3369 = 33.69% probability that 3 of them live within five miles of you.

Item e:

The expected value of the binomial distribution is:

E(X) = np

Hence, since n = 5, p = 0.55:

E(X) = 5(0.55) = 2,75

It is expected that 2.75 students live within five miles of you.

A similar problem is given at brainly.com/question/25343741

7 0
3 years ago
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