All categories

4 years ago

Write a system of equations to describe the situation below, solve using substitution, and fill in the blanks.

Mathematics

1 answer:

alexandr1967 [171]4 years ago

4 0

The total cost of the bracelet will be $96.

The number of charms will be 2.

Step-by-step explanation:

Given,

Cost per charm at Dayton Fine Jewelry = $17

Cost of bracelet = $62

Let,

x be the number of charms

y be the total cost

y = 17x+62 Eqn 1

Cost per charm at Cain Jewelers = $18

Cost of bracelet = $60

y = 18x+60 Eqn 2

For same cost;

Eqn 1 = Eqn 2

$17x+62=18x+60\\62-60=18x-17x\\2=x\\x=2$

Putting x=2 in Eqn 1

$y=17(2)+62\\y=34+62\\y=96$

Putting x=2 in Eqn 2

$y=18(2) +60\\y=36+60\\y=96$

The total cost of the bracelet will be $96.

The number of charms will be 2.

Keywords: equation, addition

Learn more about addition at:

#LearnwithBrainly

You might be interested in

Identify the number of molecules in the chemical formula for Butane.

GenaCL600 [577]

The answer is 5 molecules

3 0

4 years ago

Read 2 more answers

Combine like terms to create an equivalent expression. -5.8c+4.2-3.1+1.4c−5.8c+4.2−3.1+1.4c

nordsb [41]

Answer:

Step-by-step explanation:

-11.6+8.4-15.5+2.8

8 0

3 years ago

Read 2 more answers

a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample s

Alex

Answer:

95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].

Step-by-step explanation:

We are given that a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample standard deviation of the number of tissues per box is 97.

Firstly, the pivotal quantity for 95% confidence interval for the population variance is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2}__n_-_1$

where, $s^{2}$ = sample variance = $97^{2}$ = 9409

n = sample of boxes = 15

$\sigma^{2}$ = population variance

Here for constructing 95% confidence interval we have used chi-square test statistics.

So, 95% confidence interval for the population variance, $\sigma^{2}$ is ;

P(5.629 < $\chi^{2}__1_4$ < 26.12) = 0.95 {As the critical value of chi-square at 14

degree of freedom are 5.629 & 26.12}

P(5.629 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 26.12) = 0.95

P( $\frac{5.629 }{(n-1)s^{2} }$ < $\frac{1}{\sigma^{2} }$ < $\frac{26.12 }{(n-1)s^{2} }$ ) = 0.95

P( $\frac{(n-1)s^{2} }{26.12 }$ < $\sigma^{2}$ < $\frac{(n-1)s^{2} }{5.629 }$ ) = 0.95

95% confidence interval for $\sigma^{2}$ = [ $\frac{(n-1)s^{2} }{26.12 }$ , $\frac{(n-1)s^{2} }{5.629 }$ ]

= [ $\frac{14 \times 9409 }{26.12 }$ , $\frac{14 \times 9409 }{5.629 }$ ]

= [5043.11 , 23401.31]

Therefore, 95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].

6 0

3 years ago

Connectivity Dortmund scrubs i need an answer plz

artcher [175]

Is there any specific answer choices? i

5 0

3 years ago

Find the angle of an supplement of an angle that measures 89

Leni [432]

It'll be an acute angle, because a right angle would've been 90° but it's 89°. An obtuse angle would've been over 90°.

6 0

3 years ago