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Alisiya [41]
3 years ago
13

Help number 15 please ?

Mathematics
2 answers:
cestrela7 [59]3 years ago
4 0
Your answer:
2 Quarters: 2x25= 50 cents.
3 Dimes: 3x10= 30 cents.
All together you’d get a total of 80 cents. ($0.80)
Anton [14]3 years ago
4 0

2 quarters and 3 dimes
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List all partitions of the set. (a) (1,2) (b) [a, b,c)
Alchen [17]

Answer: Hello mate!

The partition of a set is defined as a partition of the set into a nonempty subset, where the set itself is a subset of himself, then the set is a partition of himself.

a) in this we have a set of two objects; A = (1,2) the partitions of this set are: (∅), (1), (2) and (1,2). Where (∅) is the null set.

b) Now we have a set of three objects; B = (a,b,c) the partitions of this set are: (∅), (a), (b), (c), (a,b), (a,c), (b,c), (a,b,c)

3 0
3 years ago
Write the equation of the line represented on the graph.
Marianna [84]

Answer:

y=-\frac{3}{2} x+3

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
I need help with all the ones that don't have answers.
zlopas [31]
<em>- What two consecutive odd integers have a sum of 48</em>?

23 + 25 = 48

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<em />-22 + -23

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37 + 38 = 75

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<em />25 + 27 + 29 = 81

Hope I could help! Have a good one. I believe that is all of the unanswered questions. If I missed one, let me know!<em />
7 0
3 years ago
What is the y intercept of 60x+55y=660 [with work shown]
adelina 88 [10]

Answer:

(11, 0)

Step-by-step explanation:

to find the y intercept, set y to 0.

60x + 55y = 660

60x + 55 * 0 = 660

60x = 660

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x = 11

y intercept is (11, 0)

4 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0
3 years ago
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