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Nata [24]
3 years ago
12

The three medals earned in the Olympics are the gold, silver, and bronze medals. In the 2014 Winter Olympics, the United States

had 28 total medal winners. There were two more gold medals than sivler medals, and the number of bronze medals was four less than the sum of the number of gold and silver medals. How many winners were there of each type of medal? Gold: Silver: Bronze:
Mathematics
1 answer:
marta [7]3 years ago
4 0

Answer:

Let x be the number of silver medals.

As there were two more gold medals than silver ones, gold medals are x+2

We also know that the number of bronze medals was 4 less than the sum of gold and silver, so if there are x + 2 of gold and x of silver, there are x+x+2-4 of bronze.

Now, we can do an equation, as we know there were a total of 28 medals:

x + x + 2 + x + x + 2 - 4 = 28

And we isolate x:

4x = 28

x = 28/4 = 7

There were 7 silver medals, so there were 9 gold ones (7-2) and 12 of bronze (9+7-4).

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Simplify x^2 + 7y - 4y + 9x^2.
Evgen [1.6K]

Answer:

10x^2+3y

Step-by-step explanation:

Add together the x^2 and the y's, so your equation is (x^2+9x^2)+(7y-4y)=10x^2+3y.

(hope this helps :P)

7 0
3 years ago
Find the distance between the two Points in simplest radical form
mezya [45]

Answer:

d=root under (x2 - x1)^2+(y2-y1)^2

Step-by-step explanation:

d=root under (-1-(-5)^2+(8-4)^2

=root under 16+16

=4root under 2.

3 0
3 years ago
1. Consider the chance experiment of selecting an adult at random from the sample. The number of credit cards is a discrete rand
Fantom [35]

Answer:

See the explanation below.

Step-by-step explanation:

Assuming the distribution table given:

X ( Number od Credit Cards)    Relative frequency

                   0                                    0.26

                    1                                    0.17

                   2                                    0.12

                   3                                    0.10

                   4                                    0.09

                   5                                    0.06

                   6                                    0.05

                   7                                    0.05

                   8                                    0.04

                   9                                    0.03

                   10                                   0.03

We can create the histogram for this data using the following code in R:

> x<-c(0,1,2,3,4,5,6,7,8,9,10)

> freq<-c(0.26,0.17,0.12,0.10,0.09,0.06,0.05,0.05,0.04,0.03,0.03)

> x1<-c(rep(0,26),rep(1,17), rep(2,12),rep(3,10), rep(4,9),rep(5,6),rep(6,5),rep(7,5),rep(8,4),rep(9,3),rep(10,3))

> hist(x1,breaks = x,main = "Histogram", ylab = "%", xlab = "Number of credit cards")

And we got as the result the figure attached. We see a right skewed distribution with majority of the values between 0 and 3

5 0
3 years ago
(2x^5z)^4 simplify ????????????????????
ser-zykov [4K]
Your answer should be

16x^20 z^4
6 0
4 years ago
Need helpppppp please.
Serjik [45]

t = # of hours worked as tutor

so 101-t = # of hours worked as waiter

amount earned = 9t+15(101-t)

= 9t+1515-15t

=-6t+1515

3 0
3 years ago
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