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ELEN [110]
3 years ago
10

Easy Question Topic: Volume Focus on question 10

Mathematics
2 answers:
gayaneshka [121]3 years ago
7 0

Answer:

120 cm^3;70 cm^3

Step-by-step explanation:

Volume = l * w * h

That means...

a) Volume = 12cm * 2cm * 5cm = 120 cm^3

b) Volume = 2.5cm * 8cm * 3.5cm = 70 cm^3

vovangra [49]3 years ago
4 0

Answer:

a 120 cm^3

b 70 cm^3

Step-by-step explanation:

The volume of a right rectangular prism is given by

V = l*w*h

10 a  V = 12*2*5

          =120 cm^3

10 b V = 2.5 *8 *3.5

           =70 cm^3

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3 years ago
Write an explicit formula for the geometric sequence below and use it to find the 9th term. 256, 64, 16, 4, . . .
Nastasia [14]

im doing this problem right now---

ok so a1=256 a2=64 a3=15 a4=4... find a9

the sequence divides by 1/4 every term

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replace n with 9

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7 0
2 years ago
(Based on Q1 ~ Q3) According to the Bureau of the Census, 18.1% of the U.S. population lives in the Northeast, 21.9% inn the Mid
vekshin1

Answer:

We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.

Step-by-step explanation:

The null Hypothesis: Geographical distribution of hotline callers could be the same as the U.S. population distribution

Alternative hypothesis: Geographical distribution of hotline callers could not be the same as the U.S. population distribution

The populations considered are the Midwest, South, Northeast, and west.

The number of categories, k = 4

Number of recent calls = 200

Let the number of estimated parameters that must be estimated, m = 0

The degree of freedom is given by the formula:

df = k - 1-m

df = 4 -1 - 0 = 3

Let the significance level be, α = 5% = 0.05

For  α = 0.05, and df = 3,

from the chi square distribution table, the critical value = 7.815

<u>Observed and expected frequencies of calls for each of the region:</u>

<u>Northeast</u>

Observed frequency = 39

It contains 18.1% of the US Population

The probability = 0.181

Expected frequency of call = 0.181 * 200 = 36.2

<u>Midwest</u>

Observed frequency = 55

It contains 21.9% of the US Population

The probability = 0.219

Expected frequency of call = 0.219 * 200 =43.8

<u>South</u>

Observed frequency = 60

It contains 36.7% of the US Population

The probability = 0.367

Expected frequency of call = 0.367 * 200 = 73.4

<u>West</u>

Observed frequency = 46

It contains 23.3% of the US Population

The probability = 0.233

Expected frequency of call = 0.233 * 200 = 46

x^{2} = \sum \frac{(O_{i} - E_{i})  ^{2} }{E_{i} } ,   i = 1, 2,.........k

Where O_{i} = observed frequency

E_{i} = Expected frequency

Calculate the test statistic value, x²

x^{2} = \frac{(39 - 36.2)^{2} }{36.2} + \frac{(55 - 43.8)^{2} }{43.8} + \frac{(60 - 73.4)^{2} }{73.4} + \frac{(46 - 46.6)^{2} }{46.6}

x^{2} = 5.535

Since the test statistic value, x²= 5.535 is less than the critical value = 7.815, the null hypothesis will not be rejected, i.e. it will be accepted. We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.  

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