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Fudgin [204]
3 years ago
8

Holland added three decimals that made a sum of 2. One of the decimals was 0.34. What are two other decimals Holland could have

used to make a sum of 2? Explain how you know
Mathematics
1 answer:
ddd [48]3 years ago
5 0
From this question, you can conclude:
1. There are 3 decimal
2. Decimal1 + decimal2+ decimal3= 2
3. The first decimal is 0.34

If you insert the 3rd into 2nd equation, it will become:
0.34+ decimal2+ decimal3= 2
decimal2+ decimal3= 2-0.34
decimal2+ decimal3= 1.56

If there are no rules regarding the second and third decimal, then you can use this both
1.01 + 0.55
0.78 + 0.78
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Triangles ABC and DEF are similar. The lengths of the sides of ABC are 12, 15, and 20. The
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Answer:

The length of the largest side of DEF is 40.

Step-by-step explanation:

Similar triangles' corresponding sides are proportional in length.

The smallest side of ABC is 12, and that is the corresponding side to the smallest side of DEF.

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24 / 12 = 2.

Since similar triangles' sides are corresponding, the largest side of ABC will correspond with the largest side of DEF.

Calculate the largest side length.

20 * 2 = 40.

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3 years ago
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3 years ago
1) 52 + 33=|<br> 2)<br> 26 = 42 =<br> 3) 72 - 2015<br> 4)<br> 32 x 22=
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5 0
4 years ago
The probability that the noise level of a wide-band amplifier will exceed 2 dB is 0.05 independently of every other amplifier. C
snow_lady [41]

Answer:

0.980

Step-by-step explanation:

The probability that the noise level of a wide-band amplifier will exceed 2 dB is 0.05

So, probability of success = 0.05

Probability of failure = 1-0.05=0.95

There are 12 amplifiers

We are supposed to find  the probability that at most two will exceed 2dB.

We will use binomial distribution

Formula : P(X=r)=^nC_r p^r q ^{n-r}

p = 0.05

q = 0.95

n = 12

We are supposed to find the probability that at most two will exceed 2dB.

So, P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)

P(X\leq 2)=^{12}C_0 P(0.05)^0 (0.95)^{12-0}+^{12}C_1 P(0.05)^1(0.95)^{12-1}+^{12}C_2 P(0.05)^2 (0.95)^{12-2}

P(X\leq 2)=\frac{12!}{0!(12-0)!} (0.05)^0 (0.95)^{12-0}+\frac{12!}{1!(12-1)!}(0.05)^1(0.95)^{12-1}+\frac{12!}{2!(12-2)!} (0.05)^2 (0.95)^{12-2}

P(X\leq 2)=0.980

Hence the probability that at most two will exceed 2dB is 0.980

7 0
3 years ago
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