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Orlov [11]
2 years ago
10

If 1/2 is the fraction representing the ratio of corresponding sides of two similar

Mathematics
1 answer:
Lynna [10]2 years ago
7 0

Answer:

1/4

Step-by-step explanation:

The ratio of the areas is the square of the ratio of the lengths of sides.

(1/2)² = 1/4

Answer: 1/4

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What is the value of the expression (3/7)(-2/5 •9/11)
Alex787 [66]

The answer is 54/385, or 0.14025974025974.

Hope this helps!

7 0
3 years ago
Suppouse you bought something that was priced at $6.95, and after sales tax the total bill was $7.61. What is the sale tax rate
leva [86]

Answer: sales tax of approximately .095 or 9.5%

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
3x-x-5=2(x+2) -9<br> please help
irga5000 [103]

Answer:

infinite solutions

Step-by-step explanation:

Simplify 3x-x-53x−x−5 to 2x-52x−5.

2x-5=2(x+2)-92x−5=2(x+2)−9

2 Expand.

2x-5=2x+4-92x−5=2x+4−9

3 Simplify 2x+4-92x+4−9 to 2x-52x−5.

2x-5=2x-52x−5=2x−5

4 Since both sides equal, there are infinitely many solutions.

Infinitely Many Solutions

6 0
3 years ago
Read 2 more answers
Ppppplllllllllzzzzzzz helppppppppp -7/9 - 5/10
Lyrx [107]
Simplify 5/10 to 1/2

find the LCD of both fractions and that would be 18

make the denominators the same as the LCD

Simplify, now denominators are equal

join the denominators

simplify it now (23/-18)

convert to mixed fraction 

Answer: -1 5/18.

7 0
3 years ago
Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.
lesya [120]

Answer:

0.1994 is the required probability.      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling =

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851

P(sample of 20 boxers is more than 167 pounds)

P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)

= 1 - P(z \leq 0.8438)

Calculation the value from standard normal z table, we have,  

P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0
2 years ago
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