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Lyrx [107]
2 years ago
7

A map has a scale of 1 cm = 30 km . If two cities are 11 cm apart on the map, what is the actual distance between the two cities

to the nearest tenth of a km !

Mathematics
1 answer:
HACTEHA [7]2 years ago
3 0
330 km is the correct answer.
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angel hikes at a rate of 2.4 miles per hour. If he hikes for 1/3 of an hour, how many miles will he hike?
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So Angel hikes at 2.4 miles per hour. This says that in one hour he hikes 2.4 miles. In 1/3 of an hour then he would hike 1/3 of 2.4 miles, which is 2.4/3, or 0.8.
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12 = ____% of 160? Thank you!
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7.5% of 160.

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Move decimal 2 places to the left to convert it to a percent, or multiple by 100.

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What is the sum of the geometric sequence 3,12,48... if there are 8 terms?
s344n2d4d5 [400]

Answer:

  C)  65,535

Step-by-step explanation:

You can add up the 8 terms ...

  3, 12, 48, 192, 768, 3072, 12288, 49152

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<em>Estimating</em>

Knowing the last term (49152) allows you to make the correct choice, since the sum will be more than that and less than double that.

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<em>Using the formula</em>

You know the formula for the sum of a geometric sequence is ...

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where a1 is the first term (3), r is the common ratio (4), and n is the number of terms (8).

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6 0
3 years ago
What is the circumference of a circle with a diameter of 72 millimeters?
OlgaM077 [116]
72×3.14=226.08 millimeters
3 0
2 years ago
An electronic device factory is studying the length of life of the electronic components they produced. The manager selects two
Temka [501]

Answer:

The confidence interval will be given by:

200000 \pm 44.72z, in which z is related to the confidence level.

For a confidence level of x%, z is the value in the z-table that has a pvalue of 1 - \frac{1 - z}{2}

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In the context of this problems:

It means that the sampling distributions of the sample mean of 500 components will be approximated normal, with mean 200,000 and standard deviation s = \frac{1000}{\sqrt{500}} = 44.72

To build the confidence interval:

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200000 \pm 44.72z, in which z is related to the confidence level.

For a confidence level of x%, z is the value in the z-table that has a pvalue of 1 - \frac{1 - z}{2}

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