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Nat2105 [25]
2 years ago
15

Write the following fractions in word. 3/4, 1/5, 3/4, 1 3/5.. please help me​

Mathematics
1 answer:
Alik [6]2 years ago
4 0

Answer:

"In word" as in word form?

3/4 = three fourths

1/5 = one fifth

3/4 = three fourths (again?)

1 3/5 = one and three fifths

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Steve drives at a constant speed of 59.3 mph. How far should he expect to drive in 7/8 hr?
Olenka [21]

Answer:

51.9 miles

Step-by-step explanation:

distance  = speed * time

d = 59.3 * 7/8

d = 51.8875 miles

rounded

51.9 miles

5 0
3 years ago
Use inductive reasoning to describe the pattern. then find the next two numbers in the pattern. –3, 9, –27, 81, . . .
erastovalidia [21]
Tmes -3 each times so next 2 numbers are -243 then 729
6 0
3 years ago
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The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 48,564 miles, with a standard
DerKrebs [107]

Answer:

0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 48564, \sigma = 3293, n = 281, s = \frac{3293}{\sqrt{281}} = 196.44

What is the probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct?

This is the pvalue of Z when X = 48101. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{48101 - 48564}{196.44}

Z = -2.36

Z = -2.36 has a pvalue of 0.0091

0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct

6 0
2 years ago
Pless help ineed it naww
aleksandr82 [10.1K]

Answer:

D

Step-by-step explanation:

It's the only equation that gets y correct in every one.

3 0
2 years ago
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A football team loses 4 yards on one play and 7 yards on another play. Write a sum of negative integers to represent this situat
Shtirlitz [24]

Answer:

-11 yards

Step-by-step explanation:

the first loss is represented by -4 and the second loss is represented by -7.

The sum of negative numbers will be:

(-4) + (-7) = -11

3 0
3 years ago
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