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
Alchen [17]
3 years ago
6

The corner convenience store sells candy bars at 4 for $2.91. At the price, how much would the store charge for 3 candy bars?

Mathematics
1 answer:
den301095 [7]3 years ago
6 0

Answer:

$2.1825 or simplified is $2.18.

Step-by-step explanation:

<em>0.7275 per candy bars.</em>

<em>0.7275  x 4 = $2.91.</em>

So according to that logic,

<em>0.7275  x 3 = $2.1825.</em>

<em />

You might be interested in
Can someone please help me find the answer to #19? I attached a photo of the problem
brilliants [131]

Answer:

21 packages

Step-by-step explanation:

4 ÷3/16= 4 x 16= 64/3=21 1/3 .     21 packages

hope this helps :)

4 0
3 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
Can someone please help! i have no clue what im doing
gulaghasi [49]

g(x) = -3x - 8


g(x) = 10

⇒ -3x - 8 = 10

⇒ -3x = 18

⇒ x= -6


g(-6) = 10 <==== answer is -6

7 0
3 years ago
375 of the students at Haldane Elementary School walk to school. If these
ELEN [110]

Answer:

506

Step-by-step explanation:

do the math

7 0
3 years ago
Help me pleaseee i need good grades
Sonja [21]
I gotchu one second❤️❤️❤️
7 0
3 years ago
Read 2 more answers
Other questions:
  • Xavier earns $20 a day plus a fee
    9·1 answer
  • Maridel wants to find out how many of her classmates plan to come to the next football game. There are 800 students in her schoo
    12·2 answers
  • A right triangle is removed from a rectangle to create the shaded region shown below. Find the area of the shaded region. Be sur
    9·2 answers
  • Can someone answer this it’s urgent
    12·1 answer
  • What postulate is this?<br> AAS<br> ASA<br> SAS<br> AAA
    10·2 answers
  • Help me find the volume of this shape. step-by-step explanation please!!
    13·1 answer
  • Drag the missing statements and reasons to the correct spot to complete the proof
    12·1 answer
  • According to the table below, what is the domain of the data?
    12·1 answer
  • 3.84761 to 3 decimal place
    10·1 answer
  • Solve for s. write answers as integers or as proper or improper fractions in simplest form.
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!