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Sergeu [11.5K]
2 years ago
14

AB= please help step by step

Mathematics
1 answer:
kramer2 years ago
5 0

Answer:

8x−14

Step-by-step explanation:

3x - 4 + 5x - 10

=3x+−4+5x+−10

Combine Like Terms:

=3x+−4+5x+−10

=(3x+5x)+(−4+−10)

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Substitute 9 into each equation to determine if 9 is the solution. Drag each equation into the correct column.
IRISSAK [1]

Answer:

m times 3 = 27

b plus 4 = 13

p minus 5 = 4

Those can all be solved with 9

the other can not

Step-by-step explanation:

i hope this helped

4 0
3 years ago
Swati recorded this data set, which contains an outlier.
podryga [215]

Answer:

lower quartile: 163  upper quartile: 184

Step-by-step explanation:

1. put the numbers in order from least to greatest. 97, 163, 169, 175, 184, 199

2. find the median. 172

3. find the median of the first 3 numbers to get your Q1. 163

4. find the median of the last 3 numbers to get your Q3. 184

3 0
3 years ago
Some people think it is unlucky if the 13th day of month falls on a Friday. show that in that there year (non-leap or leap) ther
Vlad1618 [11]
<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths. Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday. Jan - Friday Feb - Monday Mar - Monday Apr - Thursday May - Saturday Jun - Tuesday Jul - Thursday Aug - Sunday Sep - Wednesday Oct - Friday Nov - Monday Dec - Wednesday Now let's count how many times for each weekday, the 13th falls there. Sunday - 1 Monday - 3 Tuesday - 1 Wednesday - 2 Thursday - 2 Friday - 2 Saturday - 1 The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc. So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get: Jan - Friday Feb - Monday Mar - Tuesday Apr - Friday May - Sunday Jun - Wednesday Jul - Friday Aug - Monday Sep - Thursday Oct - Saturday Nov - Tuesday Dec - Thursday And the weekday totals are: Sunday - 1 Monday - 2 Tuesday - 2 Wednesday - 1 Thursday - 2 Friday - 3 Saturday - 1 And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year. And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
5 0
2 years ago
Pls just type it in no files pls and thank you
Nana76 [90]

Answer:

31

Step-by-step explanation:

First type of clay required 3 cups of flour per batch

She made 5 batches

5 * 3 = 15

So she used 15 cups of flour for the first type of clay

Second kind of clay required 4 cups of flour per batches

She made 4 batches

4 * 4 = 16

So she used 16 cups of flour for the second type of clay

we now add the total number of cups of flour used for each type of clay

16 + 15 = 31

She used a total of 31 cups of flour

5 0
3 years ago
A specific automotive part that a service station stocks in its inventory has an 8% chance of being defective. Suppose many cars
mylen [45]

Answer:

6.23% probability that the fourth part retrieved from stock is the first defective

Step-by-step explanation:

For each part, we have that:

8% probability of being defective.

100-8 = 92% probability of not being defective.

The parts are independent of each other.

What is the probability that the fourth part retrieved from stock is the first defective?

The first three work correctly, each one with a 92% probability.

The fourth is defective, with an 8% probability.

P = 0.92*0.92*0.92*0.08 = 0.0623

6.23% probability that the fourth part retrieved from stock is the first defective

3 0
3 years ago
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