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Brut [27]
2 years ago
12

Liquid A has a density of 0.7 g/cm³.

Mathematics
1 answer:
inn [45]2 years ago
4 0

Answer:

0.957g/cm3

Step-by-step explanation:

Density*Volume= mass

volume=mass/density

Liquid A= 140g/0.7

=200

Liquid B= 128/1.6

=1.6

Volume of liquid C= 200+80

=280

Mass of liquid C= 140+ 128

=268

Liquid C's density= mass/volume

=268/280

=0.957g/cm3

I hope I helped :)

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Area (A2) of the two surfaces wit a hole = 2(L*W - 2πd^2/4) = 2(4*4-π*2^2/4) = 25.72 cm^2
Area (A3) of  the hole = πD*W = π*2*4 = 25.13 cm^2

Total surface area, A = A1+A2+A3 = 64+25.72+25.13 = 114.85 cm^2
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3 years ago
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Svetradugi [14.3K]

Answer:

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Step-by-step explanation:


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4 years ago
Is this "N2 + 3H2 → 2NH3" balanced or no?
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3 years ago
A soft drink machine outputs a mean of 2323 ounces per cup. The machine's output is normally distributed with a standard deviati
viktelen [127]

Answer:

Area under the normal curve: 0.6915.

69.15% probability of putting less than 24 ounces in a cup.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 23, \sigma = 2

You have been asked to calculate the probability of putting less than 24 ounces in a cup.

pvalue of Z when X = 24. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{24 - 23}{2}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915

Area under the normal curve: 0.6915.

69.15% probability of putting less than 24 ounces in a cup.

5 0
3 years ago
A survey was conducted that asked 1003 1003 people how many books they had read in the past year. results indicated that x overb
kirill115 [55]
The confidence interval is

10.5\pm1.35 = (9.15, 11.85).  This means that we can be 99% confident that the mean number of books people read lies between 9.15 and 11.85.

To find the confidence interval, we first find the z-score associated with it:

Convert 99% to a decimal:  0.99
Subtract from 1:  1-0.99=0.01
Divide by 2:  0.01/2 = 0.005
Subtract from 1:  1-0.005 = 0.995

Using a z-table (http://www.z-table.com) we see that this is associated with a z-score between 2.57 and 2.58.  Since both are equally far from this value we will use 2.575.

We calculate the margin of error using

ME=z*\frac{s}{\sqrt{n}}
\\
\\=2.575*\frac{16.6}{\sqrt{1003}}=1.35

This means that the confidence interval is
10.5\pm1.35

The lower limit is given by 10.5-1.35 = 9.15.
The upper limit is given by 10.5+1.35 = 11.85
5 0
3 years ago
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