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3 years ago

Simplify the ratio 48:30.Please help me

Mathematics

2 answers:

umka21 [38]3 years ago

7 0

48 : 30 = 24 : 15

= 8 : 5

krek1111 [17]3 years ago

5 0

Answer:

don't know SORRY but I will try it

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Suppose you always eat a particular brand of cookies out of the standard sized package it comes in from the store. You notice th

Blababa [14]

Answer:

There are 91 cookies in the package of cookies

Step-by-step explanation:

The given parameter for the number of cookies are;

The number of cookies left when eating 3 at a time = 1 cookie

The number of cookies left when eating 5 at a time = 1 cookie

The number of cookies left when eating 7 cookies at a time = No cookie leftover

Let 'n' represent the number of cookies in the pack, let 'a', 'b', and 'c' be the multiples of 3s, 5s, and 7s in 'n', respectively, we have;

The number of cookies in the pack, n < 100

Therefore, we have;

3·a = n - 1

5·b = n - 1

7·c = n

∴ n - 1 is a multiples of 3 and 5 less than 100 which are;

15, 30, 45, 60, 75, and 90

Therefore, the possible values of 'n' are'

n = n - 1 + 1 = 15 + 1 = 16, 30 + 1 = 31, 45 + 1 = 46, 60 + 1 = 61, 75 + 1 = 75, 90 + 1 = 91

Multiples of 7 includes;

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, and 98

Therefore, n = 91 which is the number that appear simultaneously in both search

The number of cookies in a package = 91 cookies

3 0

3 years ago

Please Help

Anna35 [415]

Answer:

1.5 Oz because it's just a random guess hopefully I'm right

I think

7 0

2 years ago

Assume the time required to complete a product is normally distributed with a mean 3.2 hours and standard deviation .4 hours. Ho

Alex787 [66]

Answer: 3.712 hours or more

Step-by-step explanation:

Let X be the random variable that denotes the time required to complete a product.

X is normally distributed.

$X\sim N(\mu=3.2\text{ hours},\ \sigma=0.4\text{ hours} )$

Let x be the times it takes to complete a random unit in order to be in the top 10% (right tail) of the time distribution.

Then, $P(\dfrac{X-\mu}{\sigma}>\dfrac{x-\mu}{\sigma})=0.10$

$P(z>\dfrac{x-3.2}{\sigma})=0.10\ \ \ [z=\dfrac{x-\mu}{\sigma}]$

As, $P(z>1.28)=0.10$ [By z-table]

Then,

$\dfrac{x-3.2}{0.4}=1.28\\\\\Rightarrow\ x=0.4\times1.28+3.2\\\\\Rightarrow\ x=3.712$

So, it will take 3.712 hours or more to complete a random unit in order to be in the top 10% (right tail) of the time distribution.

3 0

3 years ago

I need help please!

lesantik [10]

Answer:

the correct answer is true

6 0

3 years ago

Read 2 more answers

Please help me with my math homework??

qaws [65]

-2 1/2 / 6 would be -5/12, just turn - 2 1/2 into a decimal and divide by 6 :)

4 0

3 years ago