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stellarik [79]
3 years ago
10

Solve for x. y= 2x y=20 Simplify your answer as much as possible.

Mathematics
2 answers:
Naya [18.7K]3 years ago
6 0

Answer:

20=2x

20/2=x

X=10

Step-by-step explanation:

brilliants [131]3 years ago
6 0

Answer:

x=10

Step-by-step explanation:

20 divided by 2 equals 10.

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Step-by-step explanation:

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Suppose a sample of 972972 tenth graders is drawn. Of the students sampled, 700700 read above the eighth grade level. Using the
max2010maxim [7]

Answer:

The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

972 students, 700 read above the eight grade level. We want the confidence interver for the proportion of those who read at or below the 8th grade level. 972 - 700 = 272, so n = 972, \pi = \frac{272}{972} = 0.28

85% confidence level

So \alpha = 0.15, z is the value of Z that has a pvalue of 1 - \frac{0.15}{2} = 0.925, so Z = 1.44.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 - 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.259

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 + 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.301

The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).

3 0
3 years ago
MCR3U1 Culminating 2021.pdf
11111nata11111 [884]

Answer:

(a) y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is 607,325

(ii) The population after 24 hours is 1,828,643

(c) The rate of increase of the population as a percentage per hour is 7.132%

(d) The doubling time of the population is approximately, 10.06 hours

Step-by-step explanation:

(a) The initial population of the bacteria, y₁ = a = 350,000

The time the colony grows, t = 12 hours

The final population of bacteria in the colony, y₂ = 800,000

The exponential growth model, can be written as follows;

y = a \cdot (1 + r)^t

Plugging in the values, we get;

800,000 = 350,000 \times (1 + r)^{12}

Therefore;

(1 + r)¹² = 800,000/350,000 = 16/7

12·㏑(1 + r) = ㏑(16/7)

㏑(1 + r) = (㏑(16/7))/12

r = e^((㏑(16/7))/12) - 1 ≈ 0.07132

The  model is therefore;

y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is given as follows;

y = 350,000 × (1 + 0.07132)⁸ ≈ 607,325.82

By rounding down, we have;

The population after 8 hours, y = 607,325

(ii) The population after 24 hours is given as follows;

y = 350,000 × (1 + 0.07132)²⁴ ≈ 1,828,643.92571

By rounding down, we have;

The population after 24 hours, y = 1,828,643

(c) The rate of increase of the population as a percentage per hour =  r × 100

∴   The rate of increase of the population as a percentage = 0.07132 × 100 = 7.132%

(d) The doubling time of the population is the time it takes the population to double, which is given as follows;

Initial population = y

Final population = 2·y

The doubling time of the population is therefore;

2 \cdot y = y \times (1 + 0.07132)^t

Therefore, we have;

2·y/y =2 = (1 + 0.07132)^t

t = ln2/(ln(1 + 0.07132)) ≈ 10.06

The doubling time of the population is approximately, 10.06 hours.

8 0
2 years ago
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