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3 years ago

Hiii please help i’ll give brainliest if you give a correct answer and maybe explain thankss

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

3 0

1,4,6
4 because 2:1 is equal to 4:2. 6 because 3:1 is equal to 6:2

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1) A school has 1600 students and they are going to vote as to whether they will completely convert the school completely off fo

fredd [130]

Answer:

Standard deviation of the students = 0.408

The students would you have to poll to be 95% confident of the outcome within /- 2% of the vote

= 0.408 X 1600

= 652.8

Step-by-step explanation:

Step(i):-

Given sample size 'n' = 1600

95% confidence interval of the margin error is determined by

$M.E = \frac{Z_{0.05} S.D}{\sqrt{n} }$

Level of significance = 0.05

Z₀.₀₅ = 1.96

Given Margin of error = 2% = 0.02

$M.E = \frac{Z_{0.05} S.D}{\sqrt{n} }$

$0.02 = \frac{1.96 X S.D}{\sqrt{1600} }$

0.02 X √1600 = 1.96 X S.D

$S.D = \frac{0.02X\sqrt{1600} }{1.96} = 0.408$

Standard deviation of the students = 0.408

The students would you have to poll to be 95% confident of the outcome within /- 2% of the vote

= 0.408 X 1600

= 652.8

5 0

2 years ago

Use the discriminant to determine the number and type of solutions the equation has. x2 + 8x + 10 = 0 A. no real solutions B. on

olga nikolaevna [1]

I THINK IS C.
IT HAS TWO RATIONAL SOLUTION

6 0

2 years ago

Read 2 more answers

Some one please help!

nasty-shy [4]

Answer:

The answers are 23, 9,200 and 26, 10,400

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they

Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

$n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276$

Number of ways of choosing 3 students From all males

$n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220$

The probability that all are male of choosing '3' students

$P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }$

$P(E) = \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}$

P(E) = 0.067 = 6.71%

Final answer:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0

3 years ago

Please answer me as soon as possible

adell [148]

Answer:????

Step-by-step explanation:

8 0

2 years ago