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4 years ago

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Amy and her sister are making lemonade for the school dance. Their recipe makes 2 quarts of lemonade. If a serving of lemonade i

s one cup, how many gallons of lemonade will Amy and her sister need to make in order to serve one serving of lemonade to 300 people?

1 answer:

Elden [556K]4 years ago

6 0

Answer:

17.5

Step-by-step explanation:

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If Raj randomly chooses a point in the square below, what is the probability that point is not in the circle? Assume that pi = 3

hodyreva [135]

Answer:

49.7%

Step-by-step explanation:

A cdircle is located within a square.

<u>Area of the circle</u>

Area = $\pi$ $r^{2}$ , where r = 4 units.

Area Circle = 50.3 units^2

<u>Area of the square</u>

Area = l*w or l^2 for a square, since l = w

Area = (10 units)^2

Area = 100 units^2

<u>Area in the square but outside the circle</u>

This is the difference [Square minus Circle Areas]

Square minus Circle Areas = 100 - 50.3 or <u>49.7 units^2</u>

<u>Probability</u>

The probability of picking a point in the square that is not in the circle is the ration of the two areas: <u>[Outside Circle/Square]x100%</u>

<u></u>

<u>(</u>49.7 units^2)/(100 units^2)x100% = 49.7%

<u></u>

6 0

2 years ago

Suppose that 80% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. Consider grou

Gwar [14]

Answer:

P=0.147

Step-by-step explanation:

As we know 80% of the trucks have good brakes. That means that probability the 1 randomly selected truck has good brakes is P(good brakes)=0.8 . So the probability that 1 randomly selected truck has bad brakes Q(bad brakes)=1-0.8-0.2

We have to find the probability, that at least 9 trucks from 16 have good brakes, however fewer than 12 trucks from 16 have good brakes. That actually means the the number of trucks with good brakes has to be 9, 10 or 11 trucks from 16.

We have to find the probability of each event (9, 10 or 11 trucks from 16 will pass the inspection) . To find the required probability 3 mentioned probabilitie have to be summarized.

So P(9/16 )= C16 9 * P(good brakes)^9*Q(bad brakes)^7

P(9/16 )= 16!/9!/7!*0.8^9*0.2^7= 11*13*5*16*0.8^9*0.2^7=approx 0.02

P(10/16)=16!/10!/6!*0.8^10*0.2^6=11*13*7*0.8^10*0.2^6=approx 0.007

P(11/16)=16!/11!/5!*0.8^11*0.2^5=13*21*16*0.8^11*0.2^5=approx 0.12

P(9≤x<12)=P(9/16)+P(10/16)+P(11/16)=0.02+0.007+0.12=0.147

7 0

3 years ago

Every time Liliana bakes a batch of brownies, she uses 34 cup of chocolate. If she has 128 cup(s) of chocolate remaining, how ma

pogonyaev

Answer:

3

MARK BRAINLIEST IF THIS HELPED!

i hope u have a great day.

6 0

3 years ago

Read 2 more answers

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago

If sinx=2cosx then, what is the value of sin2x?

Travka [436]

sin(x) = 2cos(x)
tan(x) = 2
tan⁻¹[tan(x)] = tan⁻¹(2)
x ≈ 63.4

sin(2x) = sin[2(63.4)]
sin(2x) = sin(126.8)
sin(2x) ≈ 0.801

3 0

3 years ago

Read 2 more answers