All categories

2 years ago

What is the greatest common factor of 12 and 18? A. 2 B. 6 C. 4 D. 12

Mathematics

2 answers:

S_A_V [24]2 years ago

4 0

Answer:

B. 6

Step-by-step explanation:

D. 12 cant be multiplied to get 18

C. 4 cant be multiplied to get 18

A. 2 is a common factor of 12 and 18 but not the greatest.

B. 6 times 2 is 12 and 6 times 3 is 18 so 6 is the greatest common factor.

ella [17]2 years ago

3 0

Answer:

Step-by-step explanation:

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 18: 1, 2, 3, 6, 9, 18

The common ones are 1, 2, 3, and 6. The greatest one among these is 6.

Factors are numbers you multiply with another natural number to get another number. For example, the factors of 16 will be all the numbers that can be multiplied with an integer to get 16. 1* 16 is 16, so 1 will work. 2 *8 is 16, so 2 will work. But, 3 * 16/3 is 16 so 3 won't work. You get the point. Hope this helps and pls give brainliset.

You might be interested in

I need help asap plz.

dangina [55]

X > 5
x = 5 is not a solution because x is always greater than 5.

3 0

3 years ago

Read 2 more answers

What + What = 69?<br> ..........................

omeli [17]

1+68
2+67
3+66
4+65
5+64
6+63
7+62
8+61
9+60

6 0

3 years ago

Of the total population of the United States, 20% live in the northeast. If 200 residents of the United States are selected at r

Snezhnost [94]

Answer:

0.0465 = 4.65% probability that at least 50 live in the northeast.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 200, p = 0.2$

$\mu = E(X) = np = 200*0.2 = 40$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.65685$

Approximate the probability that at least 50 live in the northeast.

Using continuity correction, this is $P(X \geq 50 - 0.5) = P(X \geq 49.5$ , which is 1 subtracted by the pvalue of Z when X = 49.5. So

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

$Z = \frac{49.5 - 40}{5.65685}$

$Z = 1.68$

$Z = 1.68$ has a pvalue of 0.9535

1 - 0.9535 = 0.0465

0.0465 = 4.65% probability that at least 50 live in the northeast.

3 0

3 years ago

Natalia bought 8 glitter wands for 50$

kirza4 [7]

Answer:

Is there more to the question?

4 0

3 years ago

The town of Hayward (CA) has about 50,000 registered voters. A political research firm takes a simple random sample of 500 of th

Ilia_Sergeevich [38]

Answer:

(0.0757;0.1403)

Step-by-step explanation:

1) Data given and notation

Republicans =115

Democrats=331

Independents=54

Total= n= 115+331+54=500

$\hat p_{ind}=\frac{54}{500}=0.108$

Confidence=0.98=98%

2) Formula to use

The population proportion have the following distribution

$\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the population proportion is given by this formula

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

We have the proportion of independents calculated

$\hat p_{ind}=\frac{54}{500}=0.108$

We can calculate $\alpha=1-conf=1-0.98=0.02$

And we can find $\alpha/2 =0.02/2=0.01$ , with this value we can find the critical value $z_{\alpha/2}$ using the normal distribution table, excel or a calculator.

On this case $z_{\alpha/2}=2.326$

3) Calculating the interval

And now we can calculate the interval:

$0.108 - 2.326\sqrt{\frac{0.108(1-0.108)}{500}}=0.0757$

$0.108 + 2.326\sqrt{\frac{0.108(1-0.108)}{500}}=0.1403$

So the 98% confidence interval for this case would be:

(0.0757;0.1403)

7 0

3 years ago