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Papessa [141]
3 years ago
8

An ant crawled 12 inches in 10 seconds. A second ant crawled 24 inches in 10 seconds. Calculate the speed of each ant, then comp

are the 2 by telling which one is faster.
Mathematics
1 answer:
Charra [1.4K]3 years ago
6 0

Answer:

Step-by-step explanation:divide 12/10 and 24/10. The larger answer will tell you the slower ant... I’m pretty sure

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The mean number of words per minute (WPM) read by sixth graders is 97 with a standard deviation of 19 WPM. If 75 sixth graders a
andrey2020 [161]

Answer:

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

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 are 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 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 this problem, we have that:

\mu = 97, \sigma = 19, n = 75, s = \frac{19}{\sqrt{75}} = 2.1939

What is the probability that the sample mean would be greater than 101.63 WPM?

This is 1 subtracted by the pvalue of Z when X = 101.63. So

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

By the Central Limit Theorem

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

Z = \frac{101.63 - 97}{2.1939}

Z = 2.11

Z = 2.11 has a pvalue of 0.9826.

1 - 0.9826 = 0.0174

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

5 0
3 years ago
Use the law of cosines to find the value of cos theta. round your answer to two decimal places
REY [17]

<u>Answer:</u>

The correct answer option is B. 0.21.

<u>Step-by-step explanation:</u>

We are to find the value of cos \theta using the law of cosines.

We know that the formula of the cosine rule is given by:

c ^ 2 = a ^ 2 + b ^ 2 - 2 a b cos C

So substituting the given values in the above formula:

10.9 ^ 2 = 5.8 ^ 2 + 10.5 ^ 2 - 2 (5.8) (10.5) cos C

118.81=33.64+110.25-121.8cosC

118.81 - (33.64+110.25) = -121.8cosC

118.81-143.86=-121.8cosC

cosC=\frac{-25.05}{-121.8}

cos C = 0.21

7 0
3 years ago
| x | &lt; 2 <br><br> solve the inequality and then graph the solution
anzhelika [568]

Answer:

Any number smaller than 2

Step-by-step explanation:

You would graph this by wrighting a solid, filled in circle with an arrow point toward the negative numbers, because those numbers are smaller than 2

3 0
2 years ago
For a button to fit through its button-hole the hole needs to be the size of the buttons diameter what size buttonhole is needed
Lena [83]

Answer:the size of the button hole is 2.35 cm

Step-by-step explanation:

For a button to fit through its button-hole, the hole needs to be the size of the buttons diameter.

The formula for determining the circumference of a circle is expressed as

Circumference = πd or 2πr

Where

d represents the diameter of the circle

π is a constant given as 3.14

From the information given, the circumference of the button is 7.38 cm. Therefore, the diameter of the button will be

d = circumference/π = 7.38/3.14

d = 2.35 cm

3 0
3 years ago
Select a Quadrant or axis where each ordered pair is located on a coordinate plane
valina [46]

Answer:

tq eddy porada for eddy ame

5 0
3 years ago
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