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3 years ago

What is the LCM of 6 and 18? A. 12 B. 36 C. 18 D. 6

Mathematics

2 answers:

Arisa [49]3 years ago

8 0

The lcm of 6 and 18 is C.)18

Sliva [168]3 years ago

7 0

The LCM is 18 so the answer is C

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Multiply the following complex numbers:

kozerog [31]

Answer:

Step-by-step explanation:

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2 years ago

A random sample of math majors taking an introductory statistics course were surveyed after completing the final exam. They were

quester [9]

Answer:

$E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8$

Now we can find the second moment with this formula:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4$

The variance would be given by:

$Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96$

And the deviation would be:

$Sd(X) =\sqrt{6.96}= 2.638$

Step-by-step explanation:

For this case we have the following distribution given:

X 1 2 7

P(X) 1/5 2/5 2/5

We need to begin finding the mean with this formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8$

Now we can find the second moment with this formula:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4$

The variance would be given by:

$Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96$

And the deviation would be:

$Sd(X) =\sqrt{6.96}= 2.638$

7 0

3 years ago

Hello. Is there anyone out there who can answer this for me? I truly appreciate it.

Kryger [21]

A. -1

(1,2) and (2,1)
slope = 2 -1 / 1 -2
slope = 1 / -1
slope = -1

7 0

2 years ago

Of 560 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared

frutty [35]

Answer:

a) (0.5256,0.5944)

c) Criticism is invalid

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 560

Proportion of mislabeled = 56%

$\hat{p} = 0.56$

a) 90% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64$

Putting the values, we get:

$0.56\pm 1.64(\sqrt{\dfrac{0.56(1-0.56)}{560}}) = 0.56\pm 0.0344\\\\=(0.5256,0.5944)$

b) Interpretation of confidence interval:

We are 90% confident that the true proportion of all seafood in the country that is mislabeled or misidentified is between 0.5256 and 0.5944 that is 52.56% and 59.44%.

c) Validity of criticism

Conditions for validity:

$np > 10\\n(1-p)>10$

Verification:

$560\times 0.56 = 313.6>10\\560(1-0.56) = 246.4>10$

Both the conditions are satisfied. This, the criticism is invalid.

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3 years ago

Please answer quickly! Carolyn makes and sells necklaces. The material to make each neckalce costs $3.20. She has 12 necklaces t

gayaneshka [121]

Answer:

Step-by-step explanation:

i took the test

6 0

3 years ago

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