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Phantasy [73]
2 years ago
14

What is slopes and how do you solve them?

Mathematics
1 answer:
lisov135 [29]2 years ago
4 0
You use rise over run

so say a graph is going up 2 for every 1 going across 

the slope would be 2 over 1 
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Sofia invests her money in an account paying 7% interest compounded semiannually. What is the effective annual yield on this acc
vfiekz [6]

Answer:

7.12

Step-by-step explanation:

The formula for the effective annual yield is given as:

i = ( 1 + r/m)^m - 1

Where

i = Effective Annual yield

r = interest rate = 7% = 0.07

m= compounding frequency = semi annually = 2

i = ( 1 + 0.07/2)² - 1

i = (1 + 0.035)² - 1

= 1.035² - 1

= 1.071225 - 1

= 0.071225

Converting to percentage

0.071225 × 100

= 7.1225%

Approximately to 2 decimal places = 7.12

Therefore, the annual effective yield = 7.12

4 0
3 years ago
Wirte a decimal for each of the following 9x100+2x10+3x0.1+7x0.001
borishaifa [10]

We need to perform multiplication and addition property for this problem to represent the final result in decimal form. So, the expression is,

9x100+2x10+3x0.1+7x0.001 .

Now 9 x 100= 900

2 x= 10= 20

3x0.1= 0.3

7x 0.001=0.007

Now we can all the numbers. Hence,

900+20+0.3+0.007=920.307

So, the decimal number is 930.307.

4 0
2 years ago
To see how costly it would be for other people to compete a similar cycling tour, you decide to work out two averages and compar
Damm [24]

Group A

Money spent by Person 1 = £500

Money spent by Person 2 = £600

Money spent by Person 3 = £900

Money spent by Person 4 = £450

Sum of money spent by group A = £2,450

Average money spent by group A = \frac{2,450}{4}

⇒ Average money spent by group A = £612.5

Group B

Money spent by Person 1 = £700

Money spent by Person 2 = £500

Money spent by Person 3 = £680

Money spent by Person 4 = £500

Sum of money spent by group B = £2,380

Average money spent by group B = \frac{2,380}{4}

⇒ Average money spent by group B = £595

Hence, Group B had lower average spend.

4 0
2 years ago
Type the equation for the graphbelow.Pi/3 2piy = [?] sin([ ]x)
Pie

To find the equation of the graph, what we do is to recognize some characteristics of a sine function:

- Amplitude: The amplitude of the sine function is the distance from the middle value or line running through the graph up to the highest point. In this case, the amplitude is 1.

- Period: The period of a sine function is defined as the length of one complete sine wave or one complete cycle of the curve. It can be found using the equation: P=2pi/B.

Why are these characteristics important?

Because the sine function has the following general form:

In this problem, we have A=1 and we know that the period equals 2pi/3. So,

Therefore, the equation of the graph is:

8 0
1 year ago
Choose 5 cards from a full deck of 52 cards with 13values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 kinds(spade, diamond, h
Delvig [45]

Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899

So 2899 possible ways.

4 0
3 years ago
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