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photoshop1234 [79]
2 years ago
5

PLEASE HELP ASAP A bucket contains 48 fluid ounces of water. How many quarts of water does the bucket contain? Enter your answer

as a decimal in the box. BLANK quarts
Mathematics
1 answer:
Ivan2 years ago
4 0

Answer:

1.5 Quarts

Step-by-step explanation:

A quart contains 32 fluid ounces. So divide 48 by 32. You get 1.5.

So the bucket contains 1.5 Quarts.

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Need your help ASAP please help
eduard

His average is 86 so it is below 90.  to get his average i just added all of his grades up to get 517 and then I divided 517 by 6 because thats how many grades he has and I got 86.

On the Grade row in the table it would go 83, 94, 79, 90, 96, 75 in order

The Above/Below 90 row would go -7, 4, 11, 7, -6, -3 in order

Lamont needs to get a 99 on his next test to have an average of exactly 90.


I HOPE THIS HELPED! PLEASE RATE ME AND MAKE MY ANSWER MOST BRAINLIEST!!

5 0
3 years ago
Describe the error(s) that occurred and show the correct solution.
RideAnS [48]

Answer:

-x^{2}+4x

Step-by-step explanation:

we have

(x^{2}+x)-(2x^{2} -3x)

step 1

Eliminate the parenthesis

x^{2}+x-2x^{2} +3x

The symbol of the term 3x is incorrect, must be positive instead of negative

so

-(-3x)=+3x

Group terms that contain the same variable

(x^{2}-2x^{2})+(x+3x)

Combine like terms

-x^{2}+4x

6 0
2 years ago
Money Flow  The rate of a continuous money flow starts at $1000 and increases exponentially at 5% per year for 4 years. Find the
77julia77 [94]

Answer:

Present value =  $4,122.4

Accumulated amount = $4,742

Step-by-step explanation:

Data provided in the question:

Amount at the Start of money flow = $1,000

Increase in amount is exponentially at the rate of 5% per year

Time = 4 years

Interest rate = 3.5%  compounded continuously

Now,

Accumulated Value of the money flow = 1000e^{0.05t}

The present value of the money flow = \int\limits^4_0 {1000e^{0.05t}(e^{-0.035t})} \, dt

= 1000\int\limits^4_0 {e^{0.015t}} \, dt

= 1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4

= 1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right]

= 1000 × [70.7891 - 66.6667]

= $4,122.4

Accumulated interest = e^{rt}\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t}} \, dt

= e^{0.035\times4}\times4,122.4

= $4,742

8 0
2 years ago
Find the formula for the n^th term of the sequence whose first few terms are -3,0,5,12,21,32,
skad [1K]

Answer:

77 is the ninth term.

Step-by-step explanation:

You start out with adding 3 and every time you add a new number you add the last number you added to it and add 2.

7 0
2 years ago
Match these equation balancing steps with the description of what was done in each step.
Alona [7]
Step 1: C. Step 2: A Step 3: B
3 0
2 years ago
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