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Korvikt [17]
1 year ago
6

. Isabella buys a 1.75 litre carton of apple juice. What is the largest number of 200 millilitre glasses that she can have from

the carton?
Mathematics
1 answer:
Free_Kalibri [48]1 year ago
6 0

Answer:

<em>The largest number of glasses of juice Isabella can have from the carton is 8.</em>

Step-by-step explanation:

<u>Ratios</u>

Knowing the total capacity of the carton of apple juice Ct, and the capacity of each glass Co, we can calculate how many glasses can be taken from the carton by finding the ratio:

\displaystyle n=\frac{Ct}{Co}

There are 1,000 milliliters in one liter, thus the carton has a capacity of

C t= 1.75*1,000 = 1,750 milliliters

Thus,

\displaystyle n=\frac{1,750}{200}

n=8.75

Since we cannot have 8.75 glasses, we round to the previous integer, then:

The largest number of glasses of juice Isabella can have from the carton is 8.

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Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu
Svetllana [295]

Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

a.

P(X>27)

First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of P(Z, you have to do the following conervertion:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.

c.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

d.

No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

4 0
2 years ago
In 2009, a city with a population of 796,000 had 2,388 births. What is the number of births per 10,000 in the city?
Tom [10]
796,000/2388=10,000/x
796,000x= (2388*10,000)
796,000x/796,000=23,880,000/796,000
x=30 births
7 0
2 years ago
Lester worked 12 hours last week at the grocery store and earned $93.00. If he continues to earn the same hourly pay, how many a
kykrilka [37]
93 / 12 = 7.75

so Lester earns $7.75 a hour.

if he earns $62 how many hours did he work.

so 62 = 7.75 (h)

62/ 7.75 = 8

so he worked 8 hours.
4 0
2 years ago
Read 2 more answers
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10? Im not sure though my friend just told me it was 10
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2 years ago
Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean len
Marina86 [1]

Answer:

Step-by-step explanation:

Hello!

You need to construct a 95% CI for the population mean of the length of engineering conferences.

The variable has a normal distribution.

The information given is:

n= 84

x[bar]= 3.94

δ= 1.28

The formula for the Confidence interval is:

x[bar]±Z_{1-\alpha/2}*(δ/n)

Lower bound(Lb): 3.698

Upper bound(Ub): 4.182

Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242

I hope it helps!

6 0
2 years ago
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