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Marianna [84]
3 years ago
7

a digital scale measures weight to the nearest 0.2 pound. which measurement shows an appropriate level of precison for the scale

? A. 130.4 pounds B. 125 pounds C.120 pounds D. 124.75 pounds
Mathematics
2 answers:
nadezda [96]3 years ago
6 0

Answer:

130.4 pounds

Step-by-step explanation:

The digital scale is measuring weight to the nearest 0.2 pounds, so basically the digital scale would show you the weight of an object that is a multiple of 0.2.

Talking about precision, precision tells us how accurate is the measured value or how close is the measured value to the actual value.

Here, scale is measuring to the nearest 0.2 pounds, so out of the given values the measurement that shows an appropriate level of precision for the scale is 130.4 pounds.

Alex777 [14]3 years ago
6 0
I think it would be A. 130.4 if it is measuring to the nearest 0.2 pound<span />
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jeyben [28]

Answer:

B

Step-by-step explanation:

6.5 * 24 = 156

Therefore B

4 0
2 years ago
Suppose there is a 13.9 % probability that a randomly selected person aged 40 years or older is a jogger. In​ addition, there is
Blababa [14]

Answer:

There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.

It would be unusual to randomly select a person aged 40 years or older who is male and jogs.

Step-by-step explanation:

We have these following probabilities.

A 13.9% probability that a randomly selected person aged 40 years or older is a jogger, so P(A) = 0.13.

In​ addition, there is a 15.6% probability that a randomly selected person aged 40 years or older is male comma given that he or she jogs. I am going to say that P(B) is the probability that is a male. P(B/A) is the probability that the person is a male, given that he/she jogs. So P(B/A) = 0.156

The Bayes theorem states that:

P(B/A) = \frac{P(A \cap B)}{P(A)}

In which P(A \cap B) is the probability that the person does both thigs, so, in this problem, the probability that a randomly selected person aged 40 years or older is male and jogs.

So

P(A \cap B) = P(A).P(B/A) = 0.156*0.139 = 0.217

There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.

A probability is unusual when it is smaller than 5%.

So it would be unusual to randomly select a person aged 40 years or older who is male and jogs.

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3 years ago
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Answer:

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Step-by-step explanation:

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\bf \begin{cases}&#10;v1=\ \textless \ -6,4\ \textgreater \ \\&#10;v2=\ \textless \ -3,6\ \textgreater \ \\&#10;------------\\&#10;v1\cdot v2=(-6\cdot -3)+(4\cdot 6)\\&#10;\qquad \qquad 42\\&#10;||v1||=\sqrt{(-6)^2+4^2}\\&#10;\qquad \sqrt{52}\\&#10;||v2||=\sqrt{(-3)^2+6^2}\\&#10;\qquad \sqrt{45}&#10;\end{cases}\implies \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{52}\cdot \sqrt{45}} \right)&#10;\\\\\\&#10;\measuredangle \theta =cos^{-1}\left(  \cfrac{42}{\sqrt{2340}} \right)\implies \measuredangle \theta \approx 29.74488129694^o
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3 years ago
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