The given conditions are two factors of the first number, 51 such that their product is 51 and the sum is the second number 20.
let say, the factors are x and y.

therefore, we can write, x.y = 51 ... (1) and x + y = 20 .... (2)

Putting the value of x from equation (2) in equation (1) we get,

(20 - y).y = 51

or, 20y - y² = 51

or, y² - 20y + 51 = 0

or, y = 17, 3

then x = 3,17

Therefore, The factors are 17 and 3.

Learn more about factors here:

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6 0

2 years ago

8 out of every 14 children in a school are boys. What is the ratio of boys to girls in the school?

DochEvi [55]

Answer:

8:6

Step-by-step explanation:

If 8 out of 14 students are boys then the remaining number of students must be girls

14-8= 6

There are 6 girls out of the 14

So as a ratio of boys to girls

8:6

7 0

3 years ago

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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

kow [346]

Answer:

The 80% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.149, 0.207).

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 $1 - \frac{\alpha}{2}$ .

Suppose a sample of 292 tenth graders is drawn. Of the students sampled, 240 read above the eighth grade level.

So 292 - 240 = 52 read below or at eight grade level, and that $n = 292, \pi = \frac{52}{292} = 0.178$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 - 1.28\sqrt{\frac{0.178*0.822}{292}} = 0.149$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 + 1.28\sqrt{\frac{0.178*0.822}{292}} = 0.207$

The 80% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.149, 0.207).

3 0

3 years ago