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Alisiya [41]
3 years ago
11

One little cat can eat a bag of treats in 15 minutes while another cat can eat the same bag of treats in 10 minutes. What part o

f the bag can they eat together in the given time? 1 minute.
Mathematics
2 answers:
jeka57 [31]3 years ago
8 0

Answer:

Step-by-step explanation:

The first cat eats 1/15 of a bag per minute.

The second cat eats 1/10 of a bag per minute.

Together, they eat 1/15 + 1/10 = 1/6 of a bag in one minute.

bulgar [2K]3 years ago
7 0

Answer:

Step-by-step explanation:

The first cat eats 1/15 of a bag per minute.

The second cat eats 1/10 of a bag per minute.

Together, they eat 1/15 + 1/10 = 1/6 of a bag in one minute.

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Y= -0,5x – 2<br> y= 1,5x + 8
Masja [62]

Answer:

a. ( 5 )  ( 0 ) − 2

= −2

b. ( 5 ) ( 1 ) + 8

= 13

Hope it helps

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Please help, no need to show work, tysmmmm!!
erik [133]

Answer:

translation

Step-by-step explanation:

5 0
3 years ago
1-(9-4) divided by 5<br><br> help please, i need to make sure im right.. work and answer?
SSSSS [86.1K]

1-(9-4)÷5

Do parenthesis first

(9-4)=5

1-5÷5

Division

5÷5=1

1-1=0

3 0
3 years ago
Read 2 more answers
Faye is making bead necklaces. She has 243 beads . Faye makes 51 with the same amount of beads on each necklace. How many beads
Andrej [43]

Faye has 243 beads.

She uses 51 for each necklace.


243/51 = 4.7...

243-51-51-51-51=39.

She can make 4 necklaces, with an extra of 39 beads.


8 0
3 years ago
Read 2 more answers
The amount of money spent on textbooks by a student during a semester at Norwich averages $340.00 with a standard deviation of $
dezoksy [38]

This question is incomplete

Complete Question

The amount of money spent on textbooks by a student during a semester at Norwich averages $340.00 with a standard deviation of $40.00. Find the probability that a randomly chosen student will spend a. less than $400 on textbooks in a semester b. between $400 and $460 on textbooks in a semester c. between $260 and $400 on textbooks in a semester

Answer:

a. 0.93319

b. 0.06546

c. 0.91044

Step-by-step explanation:

We solve using z score

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = $340

σ is the population standard deviation = $40

a. less than $400 on textbooks in a semester

For x = $400

z = 400 - 340/40

z = 1.5

Probability value from Z-Table:

P(x<400) = 0.93319

b. between $400 and $460 on textbooks in a semester

For x = $400

z = 400 - 340/40

z = 1.5

Probability value from Z-Table:

P(x ≤ 400) = P(x = 400)

= 0.93319

For x = $460

z = 460 - 340/40

z = 3

Probability value from Z-Table:

P(x≤ 460) = (x = 460)

= 0.99865

The probability that a randomly chosen student will spend between $400 and $460 on textbooks in a semester is

P(x = $460) - P(x = $400)

= 0.99865 - 0.93319

= 0.06546

c. between $260 and $400 on textbooks in a semester

For x = $260

z = 260 - 340/40

z = -2

Probability value from Z-Table:

P(x≤ 260) = P(x = 260)

= 0.02275

For x = $400

z = 400 - 340/40

z = 1.5

Probability value from Z-Table:

P(x ≤ 400) = P(x = 400)

= 0.93319

The probability that a randomly chosen student will spend between $260and $400 on textbooks in a semester is

P(x = $400) - P(x = $260)

= 0.93319 - 0.02275

= 0.91044

4 0
3 years ago
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