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Y_Kistochka [10]
3 years ago
8

Compare the decimals 7,231.60 and 7,231.6. Which comparison is correct? A. 7,231.60 < 7,231.6 B. 7,231.60 = 7,231.6 C. 7,231.

60 > 7,231.6\
Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
3 0
B. 7,231.60 = 7,231.6
Zinaida [17]3 years ago
3 0

the answer is  B. 7,231.60 = 7,231.6  because the zero added to the end does not count


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Which equation can be used to find the volume of the cylinder?
sergejj [24]

Answer:

V = π × 4² × 16  

Step-by-step explanation:

The formula for the volume of a cylinder is

V = πr²h

Data:

r =   4 in

h = 16 in

Calculation:

V = πr²h = π × 4² × 16

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3 years ago
Whats the answer for number 11?
viva [34]

Answer:

Option 4

Step-by-step explanation:

y = -5x + 2

-8 = -5(2) + 2

-8 = -10 + 2

Correct! :)

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2 years ago
Help me... :/<br><br> Find the measure of each exterior angle of a regular 45-gon.
GuDViN [60]

Answer:

7740° / 45

Step-by-step explanation:

The sum of the interior angles is given by this formula: (n - 2) * 180°. So a regular 45-gon would have interior angles that add up to: (45 - 2) * 180°. = 43 * 180°. = 7740°. Now divide that by the 45 angles to find the measure of one interior angle

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7 0
3 years ago
The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi
Orlov [11]

Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions 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.

In this question, we have that:

\mu = 51200, \sigma = 8200

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

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

Z = \frac{65000 - 51200}{8200}

Z = 1.68

Z = 1.68 has a pvalue of 0.954

X = 55000

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

Z = \frac{55000 - 51200}{8200}

Z = 0.46

Z = 0.46 has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

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

Z = \frac{48000 - 51200}{8200}

Z = -0.39

Z = -0.39 has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

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

Z = \frac{41000 - 51200}{8200}

Z = -1.24

Z = -1.24 has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

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

Z = \frac{61200 - 51200}{8200}

Z = 1.22

Z = 1.22 has a pvalue of 0.889

X = 41200

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

Z = \frac{41200 - 51200}{8200}

Z = -1.22

Z = -1.22 has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

4 0
2 years ago
It costs Luis 5 to park his car at a parking meter for two hours. What is the price to park for 1 hour? Draw a bar model or writ
maksim [4K]

Answer:

The cost of parking the car for 1 hour = $2.5

Step-by-step explanation:

Cost of parking car for two hours = $5

Now, let the cost of parking the car for one hour be $x

Since, the car is parked for two hours and the cost charged is $5

So, each hour the cost of parking is same ⇒ Cost of parking one hour is half the cost of parking the car for two hours.

Since, the cost of parking the car for one hour = $x

So, cost of parking the car for two hours = 2x

But this cost is given to be $5

⇒ 2x = 5

⇒ x = 2.5

Hence, The cost of parking the car for 1 hour = $2.5

3 0
3 years ago
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