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2 years ago

What is the decimal multiplier to decrease by 68%?

Mathematics

2 answers:

Brums [2.3K]2 years ago

6 0

Answer

Step-by-step explanation:

Novosadov [1.4K]2 years ago

5 0

Answer: .68

Step-by-step explanation: 68 percent in decimal form is .68 therefore you multiply a given number by .68 to decrease it by 68 percent

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A piggy bank contains some dimes and nickels. There are 8 more dimes than nickels in the bank. There is a total of $1.40. How ma

Softa [21]

Answer:

4 nickels

12 dimes

Step-by-step explanation:

Dimes are worth .1 each while nickels are .05 each.

We have 8 more dimes than nickels. Let d represent number of dimes and n for number of nickels. This means we have d=8+n.

If all our nickels and dimes together are worth 1.4 then we have another equation .1d+.05n=1.4

Lets put our equations together:

d=8+n

.1d+.05n=1.4

‐-----------------Plug first equation into second.

.1(8+n)+.05n=1.4

Distribute

.8+.1n+.05n=1.4

Combine like terms

.8+.15n=1.4

Subtract .8 on both sides

.15n=.6

Divide both sides by .15

n=.6/.15=4

Remember there are 8 more dimes so d=8+4=12.

5 0

2 years ago

The graph of the function f(x) = (x + 6)(x + 2) is shown. Which statements describe the graph? Check all that apply.

Ira Lisetskai [31]

The answers to the question would be number 2, number 3, and number 5

5 0

2 years ago

Read 2 more answers

Use the graphing calculator to locate the solutions of this system of equations:

777dan777 [17]

Hello,

y=(x-2)²-15
y=5x-1

==> 5x-1=x²-4x+4-15
==> x²-9x-10=0
Δ=9²+4*10=11²
x=(9+11)/2 or x=(9-11)/2
==>(x=10 and y=5*10-1=49) or (x=-1 and y=5*(-1)-1=-6)

1) A (-1,-6)
2) C (10,49)

7 0

2 years ago

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Player V and Player M have competed against each other many times. Historical data show that each player is equally likely to wi

Helen [10]

Answer:

0.46 (46%)

Step-by-step explanation:

We have the following data:

- Probability that player V wins the first set:

$p(1)=0.50$

(because the text says the two players are equally likely to win the first set)

- Probability that player V wins the 2nd set if he has won the 1st set:

$p(2|1)=0.60$

So, the probability that player V wins the first 2 sets is:

$p(12)=p(1)\cdot p(2|1)=(0.50)(0.60)=0.30$ (1)

Instead, the probability that player V loses the 2nd set if he has won the 1st set is 0.40 (=1-0.60), so

$p(2^c|1)=0.40$

So, the probabiity that player V winse the 1st set but loses the 2nd set is

$p(12^c)=p(1)\cdot p(2^c|1)=(0.50)(0.40)=0.20$ (2)

Also, we have:

- Probability that player V loses the 1st set:

$p(1^c)=0.50$

- Probability that she will lose the 2nd set in this case is 0.70, it means that the probability that she will win the 2nd set if she lost the 1st set is 0.30, so:

$p(2|1^c)=0.30$