D(x) = -2|x^(2 )-3.6| for x= 5.5

A card is selected at random from a deck of cards. What is the probability of randomly selecting a diamond or a 7?

Dominik [7]

Answer:

your answer is 4/13ths. #markasbrainliest

7 0

2 years ago

I need the know the answer in A B C because I’m confused

bulgar [2K]

Answer:

Step-by-step explanation:

a) edj

b) ed

c) i think ag

4 0

2 years ago

A collection of 20 coins made up of only nickels, dimes, and quarters has a total value of $3.35. If the dimes were nickels, the

Kay [80]

Answer: The required number of quarters in the collection is 11.

Step-by-step explanation: Given that a collection of 20 coins made up of only nickels, dimes and quarters has a total value of $3.35.

If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of $2.75.

We are to find the number of quarters in the collection.

Let x, y and z represents the number of nickels, dimes and quarters respectively in the collection.

We will be using the following values of nickels, dimes and quarters in form of dollar :

1 nickel = $ 0.05, 1 dime = $ 0.10 and 1 quarter = $0.25.

Then, according to the given information, we have

$x+y+z=20\\\\\Rightarrow x=20-y-z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\0.05x+0.10y+0.25z=3.35\\\\\Rightarrow 5x+10y+25z=335\\\\\Rightarrow x+2y+5z=67~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\0.05y+0.25x+0.10z=2.75\\\\\Rightarrow 5y+25x+10z=275\\\\\Rightarrow y+2z+5x=55~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of x from equation (i) in equations (ii) and (iii), we have

$(20-y-z)+2y+5z=67\\\\\Rightarrow y+4z=47\\\\\Rightarrow y=47-4z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iv)$

and

$y+2z+5(20-y-z)=55\\\\\Rightarrow -4y-3z=-45\\\\\Rightarrow y=\dfrac{45-3z}{4}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(v)$

Comparing the values of y from equations (iv) and (v), we get

$47-4z=\dfrac{45-3z}{4}\\\\\Rightarrow 188-16z=45-3z\\\\\Rightarrow 16z-3z=188-45\\\\\Rightarrow 13z=143\\\\\Rightarrow z=\dfrac{143}{13}\\\\\Rightarrow z=11.$

Thus, the required number of quarters in the collection is 11.

7 0

2 years ago

Two integers, a and b have a product of 15. what is the least possible sum of a and b?

Neporo4naja [7]

If they're integers, that means they can't be fractions (such as 15/2 and 2).

So the only couples of integers that, when multiplied, result in 15 are:

1 and 15
5 and 3.

(there are no other possibilities as both 5 and 3 are not divisible further).
The smaller of the two sums (15+1; 5+3) is 5+3=8.

The least possible sum of a and b is 8.

7 0

2 years ago

−2n < −12 <br> how can i slove this?

Ket [755]

Answer:

n>6

Step-by-step explanation:

you need to find the n variable by dividing

8 0

2 years ago

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