1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
chubhunter [2.5K]
3 years ago
10

A company sell earbuds for $17 each. At the end of the week,they have sold $731 worth of earbuds.how many earbuds did they sell?

Write and solve an equation to find the answer
Mathematics
1 answer:
Paha777 [63]3 years ago
3 0

Answer:

43 earbuds

Step-by-step explanation:

Let x be the number of earbuds sold over the week

This implies that

1 earbud:$17

x earbud:$731

Writing an equation, we get

\frac{1}{17} =  \frac{x}{731}

cross multiplying, we obtain

17x=731

Dividing through by 17,we get

x=43

You might be interested in
PLEASE HELP!! WILL MARK BRAINLEST!!!
miss Akunina [59]

Answer:

x: length

y: wide

x = 3y and x*y =192

=> 3y^2 = 192

=> y = 8 => x =3x8 = 24

=> Perimeter = (x+y)*2 = (8+24)*2 = 64

5 0
3 years ago
Read 2 more answers
Distance: (-2,7) and (2,-5)
scZoUnD [109]
Does it say anything about slope?
6 0
3 years ago
15/32 × 8/15??<br>please answer quick​
Yuliya22 [10]

Answer:

1/4

Step-by-step explanation:

3 0
3 years ago
What is the value of the expression –40 – (–50 – 2) + 10?<br><br><br> Plz help
Dafna1 [17]
The answer is 22 I looked it up to make sure i was right
7 0
3 years ago
A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the
Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample average age = 25 years

            \sigma = population standard deviation = 1.8 years

            n = sample of students = 36

            \mu = population average age

So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0
3 years ago
Other questions:
  • ♫ I have a Brainliest♫ Math situation
    10·1 answer
  • How to turn 7/50 into a decimal
    13·2 answers
  • What is 1 more than 103
    11·2 answers
  • What is the answer to 1÷⅔×(-8)×3÷(-½)
    13·2 answers
  • A penguin walks 10 feet in 6 seconds. At this speed:
    14·2 answers
  • Suppose you deposited ​$200 in a savings account 4 years ago. The simple interest rate is 2.1%. The interest that you earned in
    15·1 answer
  • !!!!!!!!!!CAN SOMEONE HELP ME WITH THESE PLEASE!!!!!!!!<br><br> FIND THE SURFACE AREA
    15·2 answers
  • HELP HELP HELP HELP HELP<br>i need clear explanation​
    5·1 answer
  • PLSS HELP MATH !!!! <br><br> 12/11 - ( -2/3 ) = ?
    5·2 answers
  • Name the postulate or theorem that proves each pair of triangles similar. Explain your reasoning.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!