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ankoles [38]
4 years ago
10

Scale drawing of the electronics board

Mathematics
1 answer:
lora16 [44]4 years ago
6 0

Area of the electronic board is 9000 cm2

Step-by-step explanation:

Write ratios to represent the unknown actual length and width of the display to the scale length and width of the display.

The Electronic board always seems like a rectangular shape

Length= scale÷Actual = 5/36 = 5×36 = 180cm

Length= scale÷Actual = 5/36 = 5×36 = 180cm

Width=scale÷Actual = 1/50 = 1×50 = 50cm

Width=scale÷Actual = 1/50 = 1×50 = 50cm

Area of Rectangle = legnth × width

                         = 180cm × 50cm = 9000 cm2

= 180cm  × 50cm = 9000 cm2

Area of the electronic board is 9000 cm2

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boyakko [2]
X => -8 -6 -8 3
y => 33 31 39 33
is not a function because the x value -8 results to two different y-values 33 and 39.
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4 years ago
To win the annual Read A Thon at his school Sebastian has to read more books than any of his classmates over winter break he sta
Mazyrski [523]

The number of sports books is 13 books.

<h3>What is an equation?</h3>

A mathematical equation is the representation of a problem by the use of variables. Often times, an equation is formed from the factors that are in a problem.

We have to represented each of the books read with an algebraic term;

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5 0
2 years ago
Can someone help me ?
antoniya [11.8K]

Answer:

AC is four times longer than DF

Step-by-step explanation:

AC is the diameter of the circle E and DE is the radius of the circle E

AC = 2 x DE

DE is the diameter of the circle F and DF is the radius of the circle F

DE = 2 x  DF

AC = 2 x DE

AC = 2 x (2 x DF)

AC = 4 x DF

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5 0
3 years ago
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Megan finds a bag of 24 craft bows at the store. The bag indicates that 23 of the bows are striped. Megan wants to know the numb
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Answer:

B

Step-by-step explanation:

4 0
3 years ago
A box contains four red balls and eight black balls. Two balls are randomly chosen from the box, and are not
castortr0y [4]

P(B) = 8/12

P(R | B) = 4/11

P(B ∩ R) = 8/33

The probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

<em><u>Solution:</u></em>

<em><u>The probability is given as:</u></em>

Probability = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}

Given that,

A box contains four red balls and eight black balls

Red = 4

Black = 8

Total number of possible outcomes = 12

Let event B be choosing a black ball first and event R be choosing a red ball second.

<h3><u>Find P(B)</u></h3>

P(B) = \frac{8}{12}

<h3><u>Find P(B n R)</u></h3>

P(B n R) = P(B) \times P(R)\\\\P(B n R) = \frac{8}{12} \times \frac{4}{11}\\\\P(B n R) = \frac{8}{33}

<h3><u>Find </u><u> P(R | B)</u></h3><h3>P(R | B) = \frac{P(R n B)}{P(B)}\\\\P(R | B) = \frac{\frac{8}{33}}{\frac{8}{12}}\\\\P(R | B) = \frac{8}{33} \times \frac{12}{8}\\\\P(R | B) = \frac{4}{11}</h3>

<em><u>The probability that the first ball chosen is black and the second ball chosen is red is about percent</u></em>

\frac{8}{33} \times 100 = 0.24 \times 100 = 24 \%

Thus the probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

4 0
4 years ago
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