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
lara [203]
3 years ago
7

Jamal finishes 5/6 of his homework.

Mathematics
1 answer:
MrRa [10]3 years ago
3 0
Jamal : 5/6
Margret : 3/4
Steve : 10/12 which reduces to 5/6

so Jamal and Steve finished the same amount
You might be interested in
At Taco Bell, I made a large purchase of 13 items that cost a total of $28.17. If Chicken Burritos cost $1.89 each and a Cheesy
Sonja [21]

Answer:

Number of Chicken Burritos purchased=7

Number of Cheesy Gordita Crunch purchased=6

Step-by-step explanation:

Let x be the number of Chicken Burritos  and y be the number of Cheesy Gordita Crunch.

Total number items purchased=13

Total cost of 13 items=$28.17

Cost of each Chicken Burritos=$1.89

Cost of each Cheesy Gordita Crunch=$2.49

We have to find number of each items purchased.

According to question

x+y=13.....(1)

1.89x+2.49y=28.17...(2)

Equation (1) multiply by 2.49 and  then subtract equation (2) from (1)

0.6x=4.2

x=\frac{4.2}{0.6}

x=7

Substitute x=7 in equation (1)

7+y=13

y=13-7=6

Number of Chicken Burritos purchased=7

Number of Cheesy Gordita Crunch purchased=6

6 0
3 years ago
Today, 32 customers at Tasha's Juice Emporium bought a total of 512 ounces of juice. How much juice did each customer buy on ave
Arturiano [62]
\frac{512 oz}{32 customers} = 16 ounces per customer
7 0
3 years ago
Read 2 more answers
There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student
Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

4 0
3 years ago
Please help me with this for brainliest <br><br><br><br> also whats yalls insta i might add you
seraphim [82]

Answer:

9

Step-by-step explanation:

divied by 3

27  divided by 3 is 9

plz mark brainliest

4 0
3 years ago
Please answer correctly !!!!!! Will mark brainliest answer !!!!!!!!!
Phoenix [80]

Answer:

see below

Step-by-step explanation:

(x + 1.5)² - 2.25 - 28 = 0

(x + 1.5)² - 30.25 = 0

(x + 1.5)² = 30.25

5 0
3 years ago
Other questions:
  • Sam and three friends shared a sandwich equally. How many equal parts were there?
    6·2 answers
  • you got 629581341 hamburgers when the restaurant seels one hundred million more what will be the answer?
    13·1 answer
  • What is F=(9/5)C+32, What is the equivalent Fahrenheit temperature of 5°C?
    11·1 answer
  • Mrs. Collins is going to plant her garden. She purchases 4 tomato plants for $2.59 each, a package of watermelon seeds for $1.87
    8·1 answer
  • Help ASAP will give brainlist
    12·1 answer
  • Which ordered pair makes both inequalities true y&gt;-3x+3 and. Ygreater than or equal to 2x-2
    14·1 answer
  • PLEASE ANSWER ILL GIVE YOU BRAINLIEST
    14·2 answers
  • 4.364923697 in radical form
    7·1 answer
  • Mr levels gave 5 tests the test scores were 200,225,215,199 and 225, what is the mean median and mode of the tests that mr level
    5·1 answer
  • I need help explaining, I know the answer, but I don't know how I got the answer
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!