Anyone know this one

here are several types of objects. For each type of object, estimate how many there are in a stack that is 5 feet high. Be prepa

lorasvet [3.4K]

We need to estimate how tall is, on average, one of these object, and then count how many would be in a 5 feet high stack.

For example, on Google you may see that "1.5 cubic foot boxes are the standard box, manufactured by most companies". So, we assume that a standard cardboard box is 1.5 feet tall.

So, if we set the equation

$1.5k = 5 \iff k=\dfrac{5}{1.5}=3.\bar{3}$

So, there would be between 3 and 4 cardboard boxes in a 5 feet tall stack.

Similarly, we can see that the average book is 9 inches tall. 9 inches are 0.75

feet, so we have

$0.75k=5 \iff k=6.\bar{6}$

So, there would be between 6 and 7 books in a 5 feet tall stack.

The average brick is 75 millimeters tall, which means 0.25 feet tall. Again, we have

$0.25k=5 \iff k=20$

So, there would be 20 bricks in a 5 feet tall stack.

Finally, a coin is about 0.006 feet, which leads to

$0.006k=5 \iff k=833.\bar{3}$

So, there would be between 833 and 834 coins in a 5 feet tall stack.

5 0

3 years ago

there are 13 animals in the barn. some are chickens and some are pigs. there are 40 legs in all. how many of each animal are the

madreJ [45]

C=chicken
p=pigs
c+p=13~p=13-c (plug in)
2c+4p=40
2c+4(13-c)=40
2c+52-4c=40
-2c=-12
c=6
6+p=13
p=7
6 chickens
7 pigs

6 0

3 years ago

Read 2 more answers

A number y minus 6 is 15

Kitty [74]

Answer:

21

Step-by-step explanation:

15 + 6= 21

so 21-6= 15

there for your answer would be 21

Mark me brainliest plz

8 0

2 years ago

a company rents out 18 food booths at the county fair. The fee for a food booth is $175 plus $7 per day. The fee for a game boot

aliya0001 [1]

Answer:

18 food booths initial costs are $(3,150 + 7d)

Step-by-step explanation:

8 0

2 years ago

The mass of a bacteria culture doubles every four days. If the initial sample is 5mg: a) how much will you have after one week (

maks197457 [2]

Answer:

a) 16.82 mg

b) 24 days

Step-by-step explanation:

Formula for exponential growth is given as:

$A=A_0b^{\frac{t}{c}}$

Where A is the final population

$A_0$ is the initial population

b is the growth factor

c is the time taken for the growth 'b'

t is the amount of time

Here, We are given the following:

$A_0$ = 5 mg

b = 2 (Given because it doubles)

c = 4 days

a) A = ?

Using the formula:

$A=5 \times 2^{\frac{7}{4}}\\\Rightarrow A=5 \times 2^{1.75} \\\Rightarrow A = \bold{16.82\ mg}$

b) Given

A = 320 mg

t = ?

Using the above formula again:

$320 = 5 \times 2^{\frac{t}{4}}\\\Rightarrow \dfrac{320} 5= 2^{\frac{t}{4}}\\\Rightarrow 2^{\frac{t}{4}} = 64\\\Rightarrow 2^{\frac{t}{4}} = 2^6\\\Rightarrow \dfrac{t}{4} = 6\\\Rightarrow t = 6 \times 4\\\Rightarrow \bold{t = 24\ days}$

So, the answers are:

a) 16.82 mg

b) 24 days

6 0

3 years ago