Plz helpp it important

Sarah’s company sells bracelets and rings. Her first customer buys 3 bracelets and 10 rings, spending a total of $48. Her second

timofeeve [1]

Sarah charges $6.00 for one bracelet.

We know that a collection of one or more linear equations containing the same variables is known as a system of linear equations in mathematics.

Let Sarah charges $x and $y for one bracelet and one ring respectively.

Since her first customer spends $48 on 3 bracelets and 10 rings, we can write it as

3x + 10y = 48 ...(1)

Since her second customer spends $54 on 7 bracelets and 4 rings, we can write it as

7x + 4y = 54 ...(2)

Now, we have to solve this system of linear equations.

Multiplying (1) by 2, 6x + 20y = 96 ...(3)

Multiplying (2) by 5, 35x + 20y = 270 ...(4)

Subtracting (3) from (4), 35x + 20y - 6x - 20y = 270 - 96

i.e. 29x = 174

i.e. x = 174/29 = 6

Therefore we can conclude that Sarah charges $6 for one bracelet. Option B is correct.

Know more about the system of linear equations here -

brainly.com/question/27664510

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3 0

2 years ago

Show work , I’ll vote you brainliest if it is correct . Thank you

kozerog [31]

Answer:

Whats the question to this problem? Do you want to know if it's correct?

i can help! Just show your work? 1/2 times whatever X is + X+1 over whatever X is = 1/2 i can get the answer if you need it unless what i said helped! :) Your welcome!

Step-by-step explanation:

Have a great week! Hope i helped! Plz dont delete answer i promise i will help and it will be correct!

6 0

2 years ago

. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$