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ankoles [38]
3 years ago
7

X is = to 1/5 of the sum of 84.7,-27.8 and 82 what is the number

Mathematics
1 answer:
Masja [62]3 years ago
4 0

Answer:

27.78

Step-by-step explanation:

x=(84.7+(-27.8)+82)÷5

x=138.9÷5

x=27.78

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8x(12+7-9)-(4+10) what’s the answer
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Equation- 8x(12+7-9)-(4+10)

Answer- 80x−14

6 0
2 years ago
Very confusing not sure
Arturiano [62]

Answer:

This describes a translation.

8 0
3 years ago
Given mſn, find the value of x.<br> (6x-7)º<br> m<br> (4x-10°
Pavel [41]

Answer:

x = 3

Step-by-step explanation:

These angles are actually equal to each other

This is because when two lines intersect, the two opposite angles are the same.

Since they are the same, you can set them equal to each other

Like so:

6x - 7 = 4x - 1

Then solve:

6x - 7 = 4x - 1

6x = 4x +6

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5 0
2 years ago
Anyone know this answer?
damaskus [11]
La of sine:

sinC/c = sinB/b==> sin  37°/8 = sin B/12 ==> sin B = 0.903


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sin(64.5) = sin(180°-64.5°) ==> B = 64.5° or 115.5°
6 0
3 years ago
The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo
muminat

Answer:

The required probability is 0.55404.

Step-by-step explanation:

Consider the provided information.

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.

Average error for 7 pages booklet and 5 pages booklet series is:

λ = 0.2×7 + 0.3×5 = 2.9

According to Poisson distribution: {\displaystyle P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}}

Where \lambda is average number of events.

The probability of more than 2 typographical errors in the two booklets in total is:

P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}

Substitute the respective values in the above formula.

P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})

P(k > 2)= 1 - (0.44596)

P(k > 2)=0.55404

Hence, the required probability is 0.55404.

4 0
3 years ago
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