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ankoles [38]
3 years ago
5

On a number line, suppose the coordinate of A is 0, and AR = 9. What are the possible coordinates of the midpoint of AR?

Mathematics
1 answer:
m_a_m_a [10]3 years ago
4 0

Given:

On a number line, suppose the coordinate of A is 0, and AR = 9.

To find:

The possible coordinates of the midpoint of AR.

Solution:

Coordinate of A is 0, and AR = 9.

It means the distance between point A and R is 9 units. It means, point R can be either left side or right side of A. So, location of point R can be either -9 or 9.

If coordinate of R is -9, then midpoint of AR is

\dfrac{0+(-9)}{2}=-4.5

If coordinate of R is 9, then midpoint of AR is

\dfrac{0+(9)}{2}=4.5

Therefore, the possible coordinates of the midpoint of AR are -4.5 and 4.5.

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Mark borrowed $5,500 at 11.5 percent for five years. What is his monthly payment?
olya-2409 [2.1K]
Use the formula of the present value of an annuity ordinary which is
Pv=pmt [(1-(1+r/k)^(-kn))÷(r/k)]
Pv present value 5500
PMT monthly payment?
R interest rate 0.115
K compounded monthly 12
N time 5years
Solve the formula for PMT
PMT=Pv÷ [(1-(1+r/k)^(-kn))÷(r/k)]
PMT=5,500÷((1−(1+0.115÷12)^(
−12×5))÷(0.115÷12))
=120.95

So the answer is C

Hope it helps!
5 0
3 years ago
Read 2 more answers
*If 3 + 4(2x - 1) = 23, then 2x - 2 equals: *​<br><br>2<br><br>4<br><br>6<br><br>8
Anarel [89]

Answer:

4

Step-by-step explanation:

3 + 4(2x - 1) = 23

4(2x - 1) = 20

8x - 4 = 20

8x = 24

x = 3

---------

2(3) - 2

6 - 2

4

7 0
3 years ago
Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,
MrRissso [65]

Answer:

4.82\cdot 10^{-4}

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

n-1=51

Therefore, the probability of drawing a three at the 2nd draw is

p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784

Then, at the third draw, the previous 2 cards are not replaced, so there are now

n-2=50

cards in the deck. So, the probability of drawing a jack is

p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08

Therefore, the total probability of drawing an ace, a three and then a jack is:

p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}

4 0
3 years ago
Help how do yall find this easy i cannot-
IgorC [24]

Answer:

153/50 or 3.06

Step-by-step explanation:

Just convert them into improper fractions and put parenthesis around the numbers with a division symbol in the middle of the two numbers.

Ex. (3/2)*(4/3)

7 0
3 years ago
Kill bought items costing $3.45,$1.99,$6.59 ,and $12.98. She used a coupon worth $2.50 .if Jill had $50.00 when she went into th
Alla [95]
First you add up all the items and the coupon then you will get $24.51. then subtract $50.00 take away $24.51 then you will get $25.49. So when she left the store she had $25.49

5 0
3 years ago
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