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LenaWriter [7]
2 years ago
11

WHATS 39 X 39 SOLVE THE PROBLEM

Mathematics
2 answers:
OleMash [197]2 years ago
7 0

Answer:

Step-by-step explanation:

= 1,521

Elis [28]2 years ago
7 0

Answer:

1521

Step-by-step explanation:

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Write an equation of a line in slope-intercept form given the slope (m) and the y- intercept (b), slope, m= -4/7 y-intercept b =
Leno4ka [110]
The answer is y=-4/7x+7. You simply substitute in the given numbers. -4/7 for the slope (m) and 7 for the y-intercept (b). 
4 0
3 years ago
Pls help …. Need it immediately ….
Leno4ka [110]

The answer is 10.73 years.

Here, the formula we can use is :

\boxed {Children = \frac{Boys+Girls}{2}} (for the average)

Substitute the values mentioned in the question.

10.54 = (10.35 + Girls)/2

21.08 = 10.35 + Girls

Girls = 10.73 years

3 0
2 years ago
At the electronics store, television screens are sized by the lengths of their diagonals. The television that Ms. Duckworth want
Inessa05 [86]

Answer:

55.8

Step-by-step explanation:

44 ^ 2 = 1936

b ^ 2 = 55 .8 ^ 2

51 ^ 2 = 2601

6 0
3 years ago
A box contains 20 light box of which five or defective it for lightbulbs or pick from the box randomly what's the probability th
Snowcat [4.5K]

Answer:

1

Step-by-step explanation:

Given:-

- The box has n = 20 light-bulbs

- The number of defective bulbs, d = 5

Find:-

what's the probability that at most two of them are defective

Solution:-

- We will pick 2 bulbs randomly from the box. We need to find the probability that at-most 2 bulbs are defective.

- We will define random variable X : The number of defective bulbs picked.

Such that,               P ( X ≤ 2 ) is required!

- We are to make a choice " selection " of no defective light bulb is picked from the 2 bulbs pulled out of the box.

- The number of ways we choose 2 bulbs such that none of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 2:

        X = 0 ,       Number of choices = 15 C r = 15C2 = 105 ways

- The probability of selecting 2 non-defective bulbs:

      P ( X = 0 ) = number of choices with no defective / Total choices

                       = 105 / 20C2 = 105 / 190

                       = 0.5526

- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:

        X = 1 ,       Number of choices = 15 C 1 * 5 C 1 = 15*5 = 75 ways

- The probability of selecting 1 defective bulbs:

      P ( X = 1 ) = number of choices with 1 defective / Total choices

                       = 75 / 20C2 = 75 / 190

                       = 0.3947

- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.

        X = 2 ,       Number of choices = 5 C 2 = 10 ways

- The probability of selecting 2 defective bulbs:

      P ( X = 2 ) = number of choices with 2 defective / Total choices

                       = 10 / 20C2 = 10 / 190

                       = 0.05263

- Hence,

    P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)

                     = 0.5526 + 0.3947 + 0.05263

                     = 1

7 0
3 years ago
How many times larger is 9x10(9) than 3x10 (-4)
Setler79 [48]
9*10(9) is equal to 810, and 3*10(-4) is equal to -120. This is 6.75 times larger than -120.
---
Hope this helps!
3 0
3 years ago
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