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EastWind [94]
3 years ago
10

PLEASE HELP I WANT TO PASS THIS LESSON PLS

Mathematics
2 answers:
zmey [24]3 years ago
4 0

Answer:

ok ok so its the first one at the top(least), then the third one, then the second one, then the fourth one at the bottom(greatest)

Step-by-step explanation:

1= 88/63 least

2= 95/18 third

3=70/27 second

4= 95/4 greatest

Arada [10]3 years ago
4 0

Answer:

Least

- 8/9 ÷ 7/11

- 8/9 ÷ 3/10

- 1 7/12 ÷ 3/10

- 1 7/12 ÷ 1/15

Greatest

Step-by-step explanation:

1 | 1.4 ≈ 8/9 ÷ 7/11

3 | 5.27 ≈ 1 7/12 ÷ 3/10

2 | 3 ≈ 8/9 ÷ 3/10

4 | 23.75 = 1 7/12 ÷ 1/15

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Suppose 40% of all college students have a computer at home and a sample of 64 is taken. What is the probability that more than
Art [367]

Answer:

0.13093

Step-by-step explanation:

Give. That :

Population mean = 40% = 0.4

Sample size (n) = 64

Probability that more than 30 have computer at home

Mean = np = 64 * 0.4 = 25.6

Standard deviation = sqrt(n*p*(1-p)) = 3.919

P(x > 30)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (30 - 25.6) / 3.919 = 1.1227353

p(Z < 1.122) = 0.13093 ( Z probability calculator)

6 0
2 years ago
PLEASE SHOW FULL SOLUTIONS !!!!! WILL MARK BRAINLIEST FOR THE BEST ANSWER. THANK YOU AND GOD BLESS
yanalaym [24]

Answer:

  • System is given below

Step-by-step explanation:

  • Let (applicable to all three lines below)
  • Hard candy = x kg with price $1.60/kg
  • Gummy worms = y kg with price $2.20/kg
  • Total weight = 50 kg with mixed price $1.75/kg

<u>Required equations:</u>

  • x + y = 50                        total weight
  • 1.60x + 2.20y = 50*1.75      total price

=========================================

<u><em>Note</em></u><em>. It says don't solve but the solution below for those who is interested to know the answer.</em>

<u>Simplify the second equation and solve by substitution x = 50 - y:</u>

  • 1.6(50 - y) + 2.2y = 87.5
  • 80 - 1.6y +2.2y = 87.5
  • 0.6y = 7.5
  • y = 7.5/0.6
  • y = 12.5

<u>Find the value of x:</u>

  • x = 50 - 12.5 = 37.5

<u>Hard candy</u> = 37.5 kg and <u>gummy worms</u> = 12.5 kg

7 0
2 years ago
Find the gof of g[5X+2]^3-4​
anygoal [31]

Answer:

what is value of f in given function

7 0
2 years ago
42/50 in simplest form
gulaghasi [49]
To find the simpliest form of a fraction, all you need to do is divide the numerator and denominator by their greatest common factor, in which in this case, is 2. And so all you need to do is divide the numerator(42) and the denominator(50) by 2, to get 21/25. This means that the simpliest form of 42/50 is 21/25.
4 0
3 years ago
Read 2 more answers
The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses.
Luden [163]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

If we had increased the confidence level, the margin of error also would have increased.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96. Increasing the confidence level, z also increases, hence the margin of error also would have increased.

The sample size and the estimate are given as follows:

n = 450, \pi = 0.47.

The lower and the upper bound of the interval are given, respectively, by:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.4239

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.5161

The 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

5 0
1 year ago
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