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WARRIOR [948]
3 years ago
6

HARDEST MATH QUESTION YOU'LL EVER SEE IN THIS WORLD...I WILL MARK BRAINLIEST AND GIVE YOU A FRIEND REQUEST. I got a screenshot o

f the problem.

Mathematics
1 answer:
yan [13]3 years ago
3 0
Each card is almost drawn at the same frequency so the answer is c. the outcomes appear to be equally likely, so a uniform probability model is a good model to represent the probabilities in Tyra's experiment. 


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If a total of 30 cups of mangoes and blueberries are used, how many batches of the recipe did the chef make
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4 years ago
(04.04)marisa painted one-half of her bedroom in three-fourths of an hour. at this rate, how long would it take her the paint th
stellarik [79]
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8 0
3 years ago
Read 2 more answers
How to find the real number solution of the equation x3+27x=9x2+27?
Alina [70]
Well think the answer is x = 14/9 but I don't think there is a full number for this equation.
3 0
3 years ago
Find the mean of the data summarized in the given frequency distribution . Compare the computed mean to the actual mean of 57.6
AlexFokin [52]

The mean of the frequency distribution is 52.6° F. The computed mean of 52.6° F is lesser than the actual mean of 57.6° F.

<h3>What is the mean of a distribution?</h3>

The mean of the distribution can be defined as the average value of the distribution, It can be expressed as the total sum of all the observed values divided by the frequency of the distribution.

From the parameters given:

Low-temperature             Frequency

40 - 44                                    2

45 - 49                                    5

50 - 54                                    9

55 - 59                                    6

60 - 64                                    3

The first thing to do is to determine the class midpoint. The class midpoint is the sum of the class interval divided by 2.

The class midpoint of 40 - 44 is \mathbf{\dfrac{40 + 44}{2} = 42}

The class midpoint of 45 - 49 is \mathbf{\dfrac{45 + 49}{2} = 47}

The class midpoint of 50 - 54 is \mathbf{\dfrac{50 + 54}{2} = 52}

The class midpoint of 55 - 59 is \mathbf{\dfrac{55 + 59}{2} = 57}

The class midpoint of 60 - 64 is \mathbf{\dfrac{60 + 64}{2} = 62}

Now, the table can be represented as:

Low-temperature             Frequency          x  

40 - 44                                    2                   42

45 - 49                                    5                   47

50 - 54                                    9                   52

55 - 59                                    6                   57

60 - 64                                    3                   62

The mean can now be determined as follows:

\mathbf{\bar x = \dfrac{\sum fx}{\sum f }}

\mathbf{\bar x = \dfrac{(2 \times 42)+ (5 \times 47) + ( 9\times 52) +(6\times 57) + (3 \times 62)}{2 + 5 + 9 + 6 + 3 }}

\mathbf{\bar x = \dfrac{1315}{25}}

\mathbf{\bar x = 52.6}

Therefore, we can conclude that the mean of the frequency distribution is 52.6° F. The computed mean of 52.6° F. is lesser than the actual mean of 57.6° F

Learn more about the mean of frequency distribution here:

brainly.com/question/12269435

3 0
2 years ago
How many dollars do you have if you have 90 nickels, 16 dimes, 16 quarters, and 7 fifty-cent pieces? Show your work and explain
tangare [24]

Answer:

$13.60

Step By Step:

A nickle is worth 5 cents.

A dime is worth 10 cents.

A quarter is worth 25 cents.

5 Cents • 90 Nickles = 450

10 Cents • 16 Dimes = 160

25 Cents • 16 Quarters = 400

50 Cents • 7 fifty-cent pieces = 350

450 + 160 + 400 + 350 = 1360

A dollar is worth 100 cents.

1360 ÷ 100 = $13.60

5 0
3 years ago
Read 2 more answers
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