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3 years ago

Unit price $12 for 150lb bag

Mathematics

1 answer:

makvit [3.9K]3 years ago

6 0

Answer:

For $1 its 12.5lbs

Step-by-step explanation:

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WILL MARK BRAINLIEST

antoniya [11.8K]

Answer:

819.54 cm^2

Step-by-step explanation:

We are given

Slant height = l = 20cm

and

Diameter = d = 18 cm

We have to find radius first

r = d/2 = 18/2 = 9 cm

The formula for surface area is:

$SA = \pi rl+\pi r^2\\= (3.14 * 9 * 20) + (3.14 * (9)^2)\\=565.5+254.34\\=819.84\ cm^2$

As second option is nearest to calculated answer, it is the correct option ..

7 0

3 years ago

Read 2 more answers

Suppose Johnny invests $2,745 into an account, which has been earning interest for many years and he now has a total of $39,974.

Doss [256]

A negative number in this wouldn't be applicable.
Johnny started out with $2,745 in the beginning, EARNING (This being the key word) interest each year. Therefore, he comes out with $39,974, not being a negative number.

Hope I helped! ^-^

5 0

3 years ago

4x - 3 < 17. Solve for x.

Andru [333]

Answer:

Step-by-step explanation:

Subtract 17 by 3, and then you get 14, multiply 4 times 3 2/3 which gives you 14 with a fraction 2/3 so the answer is c

5 0

3 years ago

In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa

Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

$^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434$

Hence, the required number of ways =27434

5 0

3 years ago

Please explain how to do this well I really need help

astraxan [27]

Simply the radical expression

6 0

3 years ago