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icang [17]
4 years ago
5

Write 47/80 as a percentage and round your answer to the nearest tenth of a percent

Mathematics
1 answer:
Amanda [17]4 years ago
8 0

Answer:

6%

Step-by-step explanation:

57 divided by 80 = 0.5875

.5875 rounded to the nearest tenth is 6%

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Problem Page
Montano1993 [528]

Answer:

425 mm


Step-by-step explanation:

The ratio of compound A to compound B is 5 is to 6.

So, we can say that the total ratio is of 11 parts (5+6=11).

<em><u>If the chemist wants a total of 935 mm of the drug, then each of the 11 parts are worth:</u></em>

\frac{935}{11}=85


<em>From the ratio, we know that Compound A is 5 parts. So:</em>

85*5=425

425 mm of Compound A is needed.

6 0
3 years ago
A number divided by three less two is at most two
Marina CMI [18]

x ≤ 12

Solution:

Let us take the number be x.

A number : x

A number divided by three : $\frac{x}{3}

A number divided by three less two : $\frac{x}{3}-2

A number divided by three less two is at most two : $\frac{x}{3}-2\leq 2

Now, simplify the inequality,

$\Rightarrow\frac{x}{3}-2\leq 2

Add 2 on both sides of the inequality,

$\Rightarrow\frac{x}{3}-2+2\leq 2+2

$\Rightarrow\frac{x}{3}\leq 4

Multiply by 3 on both sides of the equation,

$\Rightarrow\frac{x}{3}\times3\leq 4\times3

$\Rightarrow x\leq 12

Hence x ≤ 12.

5 0
4 years ago
pls solve quickly!: It took a boat 2 hours to reach town A going upstream. The way back was 1h20 min. What is the speed of the b
goldenfox [79]

Answer:

15 mph

Step-by-step explanation:

Given: Boat took 2 hours to reach Town A going upstream.

           Speed of stream= 3 mph

           Time taken to reach back home= 1 hours 20 minutes

Lets assume distance covered one side be "d" and speed of boat in still water be "s".

∴ Speed of boat in upstream= (s-3) \ mph

   Speed of boat in downstream= (s+3)\ mph

Also converting into fraction of time taken to reach back home.

Remember; 1 hour= 60 minutes

∴ Time taken to reach back home= 60+20= 80\ minutes

Converting time given into fraction= \frac{80\ minutes}{60\ minutes} = \frac{4}{3} \ hours

hence, Time taken to reach back home is \frac{4}{3} \ hours

Now forming equation of boat travelling upstream and downstream, considering distance remain constant.

We know, Distance= speed \times  time

⇒ (s-3)\times 2= (s+3)\times \frac{4}{3}

Using distributive property of multiplication

⇒2s-6= \frac{4}{3}s +4

subtracting both side by \frac{4}{3} s

⇒2s-\frac{4}{3} s-6= 4

Adding both side by 6

⇒ 2s-\frac{4}{3} s= 10

taking LCD as 3

⇒ \frac{2}{3} s= 10

Multiplying both side by \frac{3}{2}

⇒s= \frac{3}{2} \times 10

∴s= 15 mph

Hence, 15 mph is the speed of the boat in still water.

8 0
4 years ago
Solve to get the value of h
iragen [17]
The answer to the question is h=64 degrees
4 0
3 years ago
Read 2 more answers
A certain bookstore chain has two stores, one in San Francisco and one in Los Angeles. It stocks three kinds of books: hardcover
riadik2000 [5.3K]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

(a)

\left[\begin{array}{cccc}T&H&S&P\\S&600&1300&2000\\L&400&300&400\end{array}\right] = A.

In the above matrix A, the columns refers the three type of books and the rows refers the from which stores the books are been sold.

The numbers represents the corresponding sales in the month of January.

The sale is same for the 6 months.

Hence, 6A = \left[\begin{array}{cccc}T&H&S&P\\S&3600&7800&12000\\L&2400&1800&2400\end{array}\right]. This matrix 6A represents the total sales over the 6 months.

(b)

If we denote the books in stock at the starting of January by B, then

B = \left[\begin{array}{cccc}T&H&S&P\\S&1000&3000&6000\\L&1000&6000&3000\end{array}\right].

Each month, the chain restocked the stores from its warehouse by shipping 500 hardcover, 1,400 softcover, and 1,400 plastic books to San Francisco and 500 hardcover, 500 softcover, and 500 plastic books to Los Angeles.

If we represent the amount restocked books at the end of each month by another matrix C, then

C = \left[\begin{array}{cccc}T&H&S&P\\S&500&1400&1400\\L&500&500&500\end{array}\right].

This restocking will be done for 5 times before the end of June.

If there would be no sale, then the stock would be

B + 5C = \left[\begin{array}{cccc}T&H&S&P\\S&1000+2500&3000+7000&6000+7000\\L&1000+2500&6000+2500&3000+2500\end{array}\right] \\= \left[\begin{array}{cccc}T&H&S&P\\S&3500&10000&13000\\L&3500&8500&5500\end{array}\right].

Since, the total sale is given by 6A, at the end of June, the inventory in each store can be shown as following,

B+5C-6A \left[\begin{array}{cccc}T&H&S&P\\S&3500&10000&13000\\L&3500&8500&5500\end{array}\right] - \left[\begin{array}{cccc}T&H&S&P\\S&3600&7800&12000\\L&2400&1800&2400\end{array}\right] \\= \left[\begin{array}{cccc}T&H&S&P\\S&-100&2200&1000\\L&1100&6700&3100\end{array}\right]

6 0
3 years ago
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