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Art [367]
3 years ago
7

Three out of every 10 students in Mr. Chan's class bring their lunch to school. Write this ratio as a percent.

Mathematics
1 answer:
aksik [14]3 years ago
3 0
3/10 = 30%

30% of students bring their lunch.
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If a sample proportion is 0.65, which range of possible values best describes
Snezhnost [94]

The range of possible values that best describes an estimate for the population parameter is; (0.5, 0.8)

<h3>How to estimate proportion?</h3>

The range of possible values that best describes an estimate for the population parameter is; (0.5, 0.8).

When we are talking about population parameters, we also have to look at the way of estimating the population, we have to add two extremes and divide by 2.

Thus, the correct one is;

(0.5 + 0.8)/2 = 0.65

Read more about Population parameter at; brainly.com/question/2292917

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7 0
1 year ago
The perimeter of a rectangle is 156 inches. The length of the rectangle is five times its width. Find the area of the rectangle.
svlad2 [7]

Answer:

a.

Step-by-step explanation:

i think

8 0
2 years ago
A large farm tractor costs $98,424. If a farmer makes 36 equal payments, how much is each payment
mrs_skeptik [129]

Answer:

$2734

Step-by-step explanation:

$98,424 / 36 = $2734

6 0
3 years ago
What is the product of 5.06 and 2.1
Oksana_A [137]

Answer:

10.626

Step-by-step explanation:

There's multiple ways to go about it. See how

5.06 * 2.1 = (5 + 0.06)(2 + 0.1) = 5*2 + 0.06*2 + 5*0.1 + 0.06*0.1 =

= 10 + 0.12 + 0.5 + 0.006 = 10.626

Please notice how breaking up the product into smaller parts has let us make all the computations without too much hassle.

7 0
2 years ago
Read 2 more answers
Given the following winning percentages of the teams in a league (for a single year) compute the within-season standard deviatio
Masja [62]

Answer:

(b) 0.251

Step-by-step explanation:

1. Standard deviation equation:

SD=\sqrt{\frac{\sum\limits^N_i {(x_{i}-X)^{2}  } }{N} }

Where X is the mean of the data, and N the amount of data. Then, N=5

2. Estimate the Mean:

X=\frac{0.750+0.750+0.200+0.600+0.200}{5}=\frac{2.5}{5}=0.5\\

3. Caclulate Standard deviation:

SD=\sqrt{\frac{{(0.750-0.500)^{2}+(0.750-0.500)^{2}+(0.200-0.500)^{2}+(0.600-0.500)^{2}+(0.200-0.500)^{2}  } }{5} }

SD=\sqrt{\frac{0.315}{5} }=\sqrt{0.063}\\SD=0.251

7 0
3 years ago
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