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Bess [88]
3 years ago
5

9 cm + 2.8 cm = 11.8

Mathematics
2 answers:
krek1111 [17]3 years ago
6 0
It should be rounded to 2 significant figures since you cannot go to 1 significant figure if there is more than 1
So 2.8 has 2 significant figures so 11.8 should be rounded to 12
den301095 [7]3 years ago
4 0
It depends on the precision you want.

Considering the data you've given, if you want the result in cm, the numbers should be rounded to 9 and 3, thus giving a total of 12, with two significant numbers.

But taking into account the precision you've reported, looks like the decimals do matter, and in that case, the numbers should be rounded to 9.0 and 2.8, giving a total of 11.8, thus, three significative figures.
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vitfil [10]
Both of them are the first answer choice
4 0
3 years ago
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An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0
3 years ago
John read the first 114 pages of a novel, which was 3 pages less than 1/3 of the novel.
Naya [18.7K]

Answer:

[351]

Step-by-step explanation:

114 + 3 = 117. 117 x 3 = 351.

8 0
3 years ago
Solve the equation. Check your answer. If necessary, round to 3 decimal places
Lana71 [14]

Answer:

x ≈ ±20.086/√(t - 1)  

Step-by-step explanation:

ln(t - 1) + ln(x²) = 6

Recall that lnu + lnv = ln(uv). Then  

ln(t - 1) + ln(x²) = ln[(t-1)x²] = 6

Take the natural antilogarithm of each side

(t - 1)x² = e⁶

Divide each side by t - 1

x² = e⁶/(t-1)

Take the square root of each side

x = ±e³/√(t - 1)

x ≈ ±20.086/√(t - 1)

6 0
3 years ago
Forty six times y is no more than 276
Step2247 [10]

Answer:

y ≤ 6

Step-by-step explanation:

<u>Step 1:  Convert words into an expression</u>

Forty six times y is no more than 276

46 * y ≤ 276

<u>Step 2:  Solve for y</u>

46 * y / 46 ≤ 276 / 46

y ≤ 6

Answer:  y ≤ 6

4 0
3 years ago
Read 2 more answers
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