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3 years ago

Find the product of 0.34 and 0.02

Mathematics

2 answers:

Sedbober [7]3 years ago

7 0

It’s 0.68 good luck, I’m here for help

dmitriy555 [2]3 years ago

6 0

Answer:

<h2>The product is 0.68</h2>

Step-by-step explanation:

A product means we are multiplying. we are multiplying 0.34 and 0.02 which is 0.68

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Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,

vladimir1956 [14]

Answer:

(a) $C(6,0) = 1$ , $C(6,1) = 6$ , $C(6,2) = 15$ , $C(6,3) = 20$ , $C(6,4) = 15$ , $C(6,5) = 6$ , $C(6,6) = 1$ .

(b) $C(7,0) = 1$ , $C(7,1) = 7$ , $C(7,2) = 21$ , $C(7,3) = 35$ , $C(7,4) = 35$ , $C(7,5) = 21$ , $C(7,6) = 7$ , $C(7,7)=1$ .

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

$C(n,k) = \frac{n!}{k!(n-k)!}$

where the symbol $n!$ is the factorial and means

$n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n$ .

By convention 0!=1. The most important property of the factorial is $n!=(n-1)!\cdot n$ , for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

$C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1$

$C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20$

$C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1$ .

(b) The explanations to the solutions is just the calculations.

$C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1$

$C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21$

$C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35$

$C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35$

$C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21$

$C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1$

For all the calculations just recall that 4! =24 and 5!=120.

6 0

3 years ago

Two numbers totals -14, and their difference is 6. Find the two numbers

natita [175]

Answer:

The two numbers are -4 and -10.

Step-by-step explanation:

x+y=-14

x-y=6

----------

2x=-8

x=-8/2

x=-4

-4+y=-14

y=-14-(-4)=-14+4=-10

x=-4, y=-10.

4 0

3 years ago

what is the mistake in this problem and why is it wrong? 15+(√9•7)÷13 = 15+(3•7)÷3 = 15+(21)÷3 = 36+3 =12

Serggg [28]

[Edit:}

Okay! So after you have 15+(21)÷3, you have to remember PEMDAS.

PEMDAS is the order in which you solve equations.

1. Parentheses: you solve everything in the parentheses first, all while following the rules of PEMDAS

2. Exponents: after you solve the things in the parentheses, you do the exponents.

3. Then you do Multiplication or Division, solving in the order from left to right.

4. After, you do Addition or Subtraction, solving in the order from left to right.

So using PEMDAS, we'll solve 15+(21)÷3.

We do division before addition, so 21/3 is 7.

Then you add 15 to 7 and get 22 as your final answer.

Hope this helps!

7 0

3 years ago

You have $200 in a savings account. Each week for 8 weeks, you take put $18 for spending money. How much money is on your accoun

IceJOKER [234]

8 * 18 = 144

200 - 144 = 56

so $56 dollars would be left in your bank account

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MrRissso [65]

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