The figure below shows a shaded rectangular region inside a large rectangle:

Mathematics

1 answer:

pshichka [43]3 years ago

5 0

Answer:

b. 58%

Step-by-step explanation:

Calculate the area of the entire rectangle using the formula A = lw.

The lowercase "L" is for length.

"w" is for width.

The lighter square is 10 units long by 5 inches wide.

A = lw

A = (10 in)(5 in) Multiply

A = 50 in²

Calculate the area for the shaded rectangle, 7 inches by 3 inches.

A = lw

A = (7 in)(3 in) Multiply

A = 21 in²

Calculate the area for the non-shaded region by subtracting the shaded area from the total area.

50 in² - 21 in² = 29 in²

The chance that a point in the large rectangle will NOT be in the shaded region is 29/50.

Convert this fraction to decimal form by using a calculator. Divide the top number by the bottom number.

29/50 = 0.58

0.58 is in decimal form. To convert it to a percentage, multiply the number by 100.

0.58 = 58%

Therefore the probability that a point chosen inside the large rectangle is not in the shaded region is 58%.

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REALLY APPRECIATE YOUR HELP<br><br>Given log x = 3 and log y = 2, prove that x^2 -7xy =3000y

anyanavicka [17]

Answer:

$if \: log(x ) = 3 \\ log_{10}(x) = 3 \\ {10}^{3} = x \\ therfore \to \: \boxed{x = {10}^{3} } \\ \\ if \: log(y ) = 2 \\ log_{10}(y) = 2 \\ {10}^{2} = y \\ therfore \to \: \boxed{y= {10}^{2} }$

Step-by-step explanation:

$x^2 -7xy =3000y \: then \to \\ x^2 =3000y + 7xy \\ {(10 ^{3} )}^{2} = 3000(10)^{2} + 7( {10}^{3} \times {10}^{2}) \\ {10}^{6} = 3000(10)^{2} + 7( {10}^{3} \times {10}^{2}) \\ {10}^{6} = 3000(10)^{2} + 7( {10}^{5} ) \\ {10}^{6} = 300000 + 700000 \\ \boxed{{10}^{6} = 1000000} \\ hence \: the \: equation \to \: x^2 -7xy =3000y \\ \boxed{ is \: corret}$