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nikklg [1K]
2 years ago
14

A student uses the ratio of 4 oranges to 6 fluid ounces to find the number of oranges needed to make 24 fluid ounces of juice. T

he student writes this proportion: StartFraction 4 over 6 EndFraction = StartFraction 24 over 16 EndFraction Explain the error in the student’s work.
Mathematics
2 answers:
Andreas93 [3]2 years ago
8 1
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Guest
1 year ago
That's not an answer.
Lorico [155]2 years ago
9 0

Answer:

16 oranges

Step-by-step explanation:

Here you go! 4 oranges is 6 fluid ounces so, 6 x 4 = 24 so you need for more bunches that have 4 in them. Therefore you must say that 4 x 4 = 16

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The following items are available for use in a pan balance to measure the weight of a large rock. Which items will yield the gre
weqwewe [10]
The correct option is B.
Pan balance is one of the instruments that are used to measure weights of objects. The instrument is designed in such a way that, it has two pans and a standard weight can be added to one of the pan to determine the weight of the object in the other pan. In the question given above, the pan balance is to be used to measure the weight of a large rock. Salt is the right object to be used on the second pan because it can be packaged in standard weights and it is very light in weight. So, it will be very sensitive to the weight of the rock that is been added on the other pan. 
8 0
2 years ago
Seams Personal advertises on its website that 95% of customer orders are received within four working days. They performed an au
Bezzdna [24]

Answer:

a. Yes(n=500>=5, n(1-p)=25>=5)

b. 0.15241

Step-by-step explanation:

a. A normal approximation to the binomial can be used  n\geq5 and n(1-p)>=5:

#We calculate our p as follows:

\hat p=x/n=470/500=0.94

n=500

n(1-p)=500(1-0.95)=25

Hence, we can use the normal approximation.

b. This is a normal approximation.

-Given that p=0.95(95%)

-We verify if our distribution can be approximated to a normal:

np=0.95\times 500=475\\n(1-p)=500(1-0.95)=25\\\\np\geq 5,\ n(1-p)\geq 5

Hence, we can use the normal approximation of the form:

P_{bin}(k,n,p)->N(\mu,\sigma^2)\left \{ {{\mu=np=475} \atop {\sigma=\sqrt{np(1-p)}=4.8734}} \right. \\\\\\P_{bin}(k\leq 470)\approx P_{norm}(x\leq 470.5)=P_{norm}(z\leq \frac{470-475}{4.8734})\\\\P_{norm}(z\leq -1.0260)=0.15241

Hence, the probability of the sample proportion  is the same as the proportion of the sample found is 0.15241

3 0
2 years ago
the expression below gives the cost of green beans(x) and cucumbers (y). what is the cause of 6 bunches of green beans and 4 cuc
kipiarov [429]

This is my explanation thus sucks and half the timem you guys dont even have the right answer . So this is my answer for you at leatst try to get it tight because if you dont try you wont suceed

3 0
2 years ago
Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum
dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

c.

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

=.

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

e.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

8 0
2 years ago
Mikayla is training for a half-marathon. At the beginning of her training, she is able to run 2 miles without stopping. Her goal
ivanzaharov [21]

Alright, lets get started.

At the beginning of her training, she is able to run 2 miles without stopping means 2 will be in the addition when we will write the equation for her running miles.

Her goal is to increase her distance by three quarters of a mile each week.

Means in first week, she will add the distance = \frac{3}{4} miles

Suppose she runs x weeks, so shee will add distance = \frac{3}{4}x

So, suppose she runs total number of miles = d, then

d = 2 + \frac{3}{4}x

Where d is total numer of miles she runs

x is number of weeks    :   Answer

Hope it will help :)


5 0
2 years ago
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