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rodikova [14]
3 years ago
15

Express 1.50 as a percentage of 2.50

Mathematics
1 answer:
LekaFEV [45]3 years ago
7 0
1.50 of 2.50 expressed as a percentage is 60%
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State 4 solutions for the equation that is modeled by the graph.
Norma-Jean [14]

Answer:

see explanation

Step-by-step explanation:

Any coordinate point that lies on the line is a solution , that is

(- 2, 8 ) , (0, 4 ) , (2, 0 ) , (4, - 4 )

7 0
3 years ago
Solve (i) 4/15+x=17/30​
Anna007 [38]

Answer:

3/10

Step-by-step explanation:

4/15+x=17/30

x=17/30-4/15

x=17/30-8/30

x=9/30

simplify

x=3/10

3 0
3 years ago
A square box has an area of 90,000 square inches. What's the length of each side of the box?
Sever21 [200]

Answer: A 300 inches

Step-by-step explanation:

Area (a) = 90,000

Side = √a

Side = √90000

Side = 300

8 0
4 years ago
Read 2 more answers
General Electric Company has an assembly line where light bulbs are produced. The ratio of defective bulbs to good bulbs produce
kati45 [8]

Answer: 200 bulbs will not be defective.

Step-by-step explanation:

The ratio of defective bulbs to good bulbs produced each day is 2 to 10. This ratio can also be expressed as 1 to 5 by reducing to lowest terms.

The total ratio is the sum of the proportions.

Total ratio = 1 + 5 = 6

This means that if n bulbs is produced, the number of defective bulbs would be

1/6 × n

The number of non defective would be

5/6 × n

Since n = 240, then the number of bulbs that will not be defective is

5/6 × 240 = 200 bulbs

4 0
4 years ago
Assume we need to estimate the mean of a normally-distributed population with great accuracy. Specifically, for significance lev
bulgar [2K]

Answer:

n = 2662.56\sigma^2                                

Step-by-step explanation:

We are given the following in the question:

Significance level = 0.01

Width of interval = 0.1

Population variance = \sigma^2

We have to find the sample size so that the width of the confidence interval is no larger than 0.1

z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58

Formula for sample size:

n = \displaystyle\frac{z^2\sigma^2}{E^2}

where E is the margin of error. Since the confidence interval width is 0.1,

E = 0.05

Putting these values in the equation:

n = \displaystyle\frac{(2.58)^2\sigma^2}{(0.05)^2} = 2662.56\sigma^2

So, the above expression helps us to calculate the sample size so that the width of the confidence interval is no larger than 0.1 for different sample variances.

4 0
4 years ago
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