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kati45 [8]
3 years ago
5

Write a proportion for each situation, Then solve

Mathematics
1 answer:
natulia [17]3 years ago
3 0

3/4: 2 , ?:7, 5 1/4:7
It takes 5 1/4 of honey to make 7 banana breads.
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GUYSSS PLSSS HELP ME WITH THIS QUESTION :))
Mrac [35]

Answer:

17

Step-by-step explanation:

1) if to solve the first inequation, tnen 3x<52; ⇔ x<52/3;

2) if to solve the seconde inequation, then 2x≥24; ⇔ x≥12.

3) according to the items 1 and 2 x∈[12;52/3);

4) the largest prime number is 17.

3 0
3 years ago
Q3. If I = (P × R × T)/100, find T if I = 80 Rs, P = 400 Rs., R = 10 Rs. (2)
sukhopar [10]

9514 1404 393

Answer:

  T = 2

Step-by-step explanation:

Fill in the given numbers and solve for T.

  I = PRT/100

  80 = 400·10·T/100

  80 = 40T . . . . . . . . . . . . simplify

  80/40 = T = 2 . . . . . . . . divide by the coefficient of T

4 0
3 years ago
Read 2 more answers
A + (b-4) when a-24 and b-7
lara31 [8.8K]

Answer:

a=24

b=7

now

24+(7-4)

24+3

27

so the answer is 27

5 0
3 years ago
Ms. Good has a bag of 15 marbles. Three are blue, six are red, and six are yellow. What is the probability that she picks a yell
Delicious77 [7]

Answer:

the chances of her picking a yellow one is 6 in 15

Step-by-step explanation:

all you have to do is divide the color that you are trying to find the probability of by the amount of that object is if that makes sense.

5 0
3 years ago
Read 2 more answers
What is a three digit number thats an odd multiple of three and the product of its digits is 24 and is larger than 15 squared?
Reika [66]
The only way 3 digits can have product 24 is 
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4 
To be divisible by 3 the sum of the digits must be divisible by 3.
1+  3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3. 

So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.

Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.

The others must end in 3. 

They must be greater than 152 which is 225. So the

First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.

The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.  
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
7 0
3 years ago
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