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lawyer [7]
3 years ago
10

(HELP FINAL EXAM) If two states are selected at random from a group of 40 ​states, determine the number of possible outcomes if

the group of states are selected with replacement or without replacement.
Mathematics
1 answer:
Hitman42 [59]3 years ago
7 0

Answer:

With replacement, there are 1600 possible outcomes.

Without replacement, there are 1560 possible outcomes.

Step-by-step explanation:

Chosen with replacement:

State is chosen, then put back in the group, then another state can be chosen. In both cases, there will be 40 possible outcomes, as the first taken state is placed back. So

40*40 = 1600

With replacement, there are 1600 possible outcomes.

Chosen without replacement:

State and chosen and is not replaced.

So, for the second state, there will be only 39 possible outcomes.

40*39 = 1560

Without replacement, there are 1560 possible outcomes.

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6.5yd and 11yd<br> please find the area
Afina-wow [57]
Multiply length times width to find the area. 6.5x11= 71.5
3 0
3 years ago
Suppose a population of 140 crickets doubles in size every month. The function f(x) = 140∗2x gives the population after x months
Tema [17]
The equation would then be f(x) = 140 x 2(24)

There will be 6,720 crickets in 2 years
8 0
3 years ago
Noel has rowing lessons every 5 days and guitar lessons every 6 days. If he had both lessons on the last day of the previous mon
bixtya [17]

Answer:

Since every 30 days  he  wil have both lessons on the same day , and  he already  had both lessons on the last day of the previous month, that means that the day 30  the current month   he  wil have both lessons on the same day (It may be the last day if the month has 30 days or it may not be the last day if the month has 31 days)

Step-by-step explanation:

Lets find the least common factor of 5 and 6

Multiples of 5

5  10  15  20  35  30  35  40......

Multiples of 6

6  12  18  24  30 36  

LCF of 5 and 6 = 30

Every 30 days  he  wil have both lessons on the same day

3 0
3 years ago
The model represents x2 – 9x + 14. An algebra tile configuration showing only the Product spot. 24 tiles are in the Product spot
klio [65]

I didn't get all the part with the tiles, but here's the general answer:

given a polynomial

p(x)=ax^2+bx+c

we have that x-k is a factor of p(x) if and only if k is a root of p(x), i.e. if

p(k)=ak^2+bk+c=0

So, given the polynomial

p(x)=x^2-9x+14

We can check if x-9 is a factor by evaluating p(9):

p(9)=81-81+14=14\neq 0

So, x-9 is not a factor.

Similarly, we can evaluate p(2),\ p(-5),\ p(-7) to check if x-2,\ x+5,\ x+7 are factors:

p(2)=4-18+14=0,\quad p(-5)=25+45+14=84\neq 0,\quad p(-7)=49+63+14=126 \neq 0

So, only x-2 is a factor of x^2-9x+14

4 0
3 years ago
Read 2 more answers
the length of a rectangular patio is 8 feet less than twice its width. the area of the patio is 280 square feet. find the dimens
jolli1 [7]

Answer:

The length of the rectangle 'l' = 20

The width of the rectangle 'w' = 14

Step-by-step explanation:

<u>Explanation</u>:-

Let 'x' be the width

Given data the length of a rectangular patio is 8 feet less than twice its width

2x-8 = length

The area of rectangle = length X width

Given area of rectangle = 280 square feet

x(2x-8) = 280

2(x)(x-4) =280

x(x-4) =140

x^2 -4x -140=0

x^2-14x+10x-140=0

x(x-14)+10(x-14)=0

(x+10)(x-14) =0

x = -10 and x = 14

we can choose only x =14

The width of the rectangle 14

The length of the rectangle 2x-8 = 2(14)-8 = 28 -8 =20

The length of the rectangle 'l' = 20

The width of the rectangle 'w' = 14

6 0
3 years ago
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